1. Solutions

What you will learn

• Classify solutions by the states of their components.
• Choose and convert concentration measures.
• Explain saturation and apply the stated Henry-law convention.
• Use Raoult’s law and distinguish liquid from vapour composition.
• Recognise ideality, deviations and azeotropes.
• Calculate colligative effects and infer molar mass.
• Interpret particle changes through a clearly stated van ’t Hoff model.

Big idea: a solution can change what a solvent does without changing what the solvent is.

Imagine two sealed containers with the same amount of water. One also holds a dissolved, nonvolatile substance. They need not have the same vapour pressure or freezing point. The difference begins with how many particles share the liquid, how they interact, and how easily solvent molecules can move into another phase.

This chapter separates three tasks. First describe composition. Next decide which equilibrium model applies. Finally calculate a property with units that fit that model. Keeping these tasks separate prevents most concentration errors. A formula is useful only after its assumptions have been identified.

During the nineteenth century, Henry and Raoult established different limiting pressure–composition relations. Van ’t Hoff connected dilute-solution particle effects with relations resembling those for gases. These names label useful models with different composition ranges; they are not interchangeable laws.

A liquid can look uniform even when it contains several substances. The useful question is whether every small sample has the same composition.

A solution is a homogeneous mixture. Its components are mixed on a molecular or ionic scale. The solvent sets the physical state of the solution and is usually the larger component. A solute is another component dissolved in it. These words describe roles in a mixture; they are not permanent labels for substances.

In a binary solution there are two components. A gas, liquid or solid can act as solute. The solvent can also have any of these states. A solution need not be colourless. Colour does not decide whether it is homogeneous.

A suspension is different: particles are large enough to form a separate dispersed phase and may settle. An emulsion contains droplets of one liquid in another. A temporary uniform-looking shake does not make these systems true solutions. State the scale and the number of phases, not just what the container looks like.

A sealed sample contains a uniform mixture of two gases. Which substance is the solvent?

For a gas–gas mixture, the larger component is normally called the solvent. The question does not give their amounts, so it does not determine that label.

Result: Both are solution components; more composition information is needed to assign the usual solvent label.

1. Is a clear but coloured salt solution automatically heterogeneous?
2. What makes a solution binary?

Answers

1. No. Colour and homogeneity are different properties.
2. It contains exactly two components.

A concentration is a comparison. Before calculating, identify both the amount in the numerator and the amount in the denominator.

Mass percentage compares the mass of one component with the mass of the complete solution. Volume percentage compares the component volume used to prepare a solution with the final solution volume. Mass-by-volume percentage means grams of solute per hundred millilitres of solution; it is not a mass fraction.

$$\text{mass percent}=\frac{m_{\rm solute}}{m_{\rm solution}}\times100$$

For a very dilute mixture, parts per million compares matching quantities and multiplies the fraction by a million. State whether the fraction is mass/mass, volume/volume or another defined basis. A ppm value without its basis can be ambiguous.

$$x_i=\frac{n_i}{\sum_jn_j},\qquad c=\frac{n_{\rm solute}}{V_{\rm solution}},\qquad m=\frac{n_{\rm solute}}{m_{\rm solvent}\;(\mathrm{kg})}$$

Mole fraction has no unit, and the mole fractions add to one. Molarity uses litres of solution. Molality uses kilograms of solvent. Heating changes volume, so molarity may change even when no material is lost. Molality and mole fraction stay the same if the component amounts stay the same.

A solution contains $11.9\,\mathrm g$ of KBr in a final volume of $0.400\,\mathrm L$. Use $M(\mathrm{KBr})=119\,\mathrm{g\,mol^{-1}}$. Find its molarity.

$$n=\frac{11.9}{119}=0.100\,\mathrm{mol}$$

Use the final solution volume, not the volume of water poured in.

Result: $c=0.25\,\mathrm{mol\,L^{-1}}$.

Dissolve $9.0\,\mathrm g$ of glucose in $90.0\,\mathrm g$ of water. Use molar masses $180$ and $18\,\mathrm{g\,mol^{-1}}$. Find three concentration measures.

The solute amount is $9/180\,\mathrm{mol}$ and the water amount is $90/18\,\mathrm{mol}$.

The solution mass is the sum of both masses. The solvent mass in the molality calculation is $0.090\,\mathrm{kg}$.

Result: Mass percentage $=9.0909\%$, molality $=0.5556\,\mathrm{mol\,kg^{-1}}$, and glucose mole fraction $=0.009901$.

Dilution adds solvent while preserving the amount of dissolved solute. For a dilution without reaction or loss, the initial concentration times the initial solution volume equals the final concentration times the final solution volume. This relation requires both concentrations to be on the same volume basis.

$$c_1V_1=c_2V_2$$

1. Dilute $40.0\,\mathrm{mL}$ of a $0.250\,\mathrm M$ solution to $200.0\,\mathrm{mL}$. Find the new molarity.
2. Which measure is most convenient for freezing-point calculations: molarity or molality?

Answers

1. $0.05\,\mathrm M$.
2. Molality, because the colligative relation is defined using the solvent mass.

Stirring can make a solid dissolve faster, but it does not necessarily change how much will dissolve at equilibrium.

Solubility is the equilibrium amount of solute that dissolves in a stated amount of solvent at a stated temperature. In a saturated solution in contact with undissolved solid, dissolution and crystallisation continue at equal rates. The solid is not motionless at molecular scale; there is no net transfer.

A useful guide is similarity of intermolecular interactions. Polar or ionic solutes often dissolve better in polar solvents, while nonpolar substances often prefer nonpolar solvents. The guide is not an absolute law. Both energy changes and the change in the number of accessible arrangements contribute to whether dissolution is favourable.

If dissolution is endothermic, warming commonly increases solubility. If it is exothermic, warming may decrease it. Use the enthalpy sign for the particular substance. Pressure has little effect on the solubility of most solids in liquids because their volumes change only slightly.

For a specified gas–solvent pair at a fixed temperature, Henry’s law relates the partial pressure of the gas above the solution to its dissolved mole fraction. Use the gas partial pressure, not the total pressure of an unrelated gas mixture.

$$p=K_Hx$$

In this convention a larger Henry constant means a smaller dissolved mole fraction at the same pressure. Do not transfer that interpretation to a different convention without checking its equation. Gas solubility often falls on warming because dissolution of many gases is exothermic.

At a fixed temperature, a gas has $K_H=8.0\times10^4\,\mathrm{bar}$. Its partial pressure is $4.0\,\mathrm{bar}$. Find its mole fraction.

Rearrange the stated convention to $x=p/K_H$. Use the same pressure unit in both quantities.

Result: $x=0.00005$. If pressure doubles while the law remains valid, the predicted mole fraction doubles.

A pressurised gas–liquid system illustrates the law: releasing the pressure can make the liquid temporarily supersaturated, so bubbles form. Henry’s law is most useful for dilute solutions without a strong chemical reaction between the dissolved gas and solvent. It is an equilibrium relation, not a rule for the speed of bubble formation.

1. A gas has $x=0.00020$ under $2.0\,\mathrm{bar}$. Predict $x$ under $5.0\,\mathrm{bar}$ at the same temperature.
2. Why is “warming always improves dissolution” incorrect?

Answers

1. $x=0.0005$, within the dilute Henry-law range.
2. The equilibrium response depends on the enthalpy of dissolution, and warming often reduces gas solubility.

A sealed liquid does not stop evaporating. Its equilibrium vapour pressure is reached when escape into the vapour and return to the liquid balance.

In an ideal liquid mixture, the partial vapour pressure of each volatile component equals its liquid mole fraction times its pure-liquid vapour pressure at the same temperature. The total vapour pressure is the sum of these partial pressures.

$$p_A=x_Ap_A^*,\quad p_B=x_Bp_B^*,\quad p_{\rm total}=x_Ap_A^*+(1-x_A)p_B^*$$

The vapour composition is generally different from the liquid composition. In the ideal-gas approximation, the vapour mole fraction of a component is its partial pressure divided by the total pressure. The more volatile component is the one with the higher pure-liquid vapour pressure at that temperature.

For an ideal mixture, $p_A^*=80\,\mathrm{kPa}$, $p_B^*=30\,\mathrm{kPa}$ and $x_A=0.400$. Find the total pressure and the vapour mole fraction of A.

Find each partial pressure separately, then add. The vapour fraction uses pressure, not the original liquid fraction.

Result: $p_A=32\,\mathrm{kPa}$, $p_B=18\,\mathrm{kPa}$, $p_{\rm total}=50\,\mathrm{kPa}$ and $y_A=0.64$.

Both relations can look like pressure proportional to mole fraction. Raoult’s law describes the solvent near its pure-liquid limit; Henry’s law describes a solute near infinite dilution. If a component obeys Raoult’s law throughout the composition range, the proportionality constant is its pure vapour pressure. Do not assume every real gas–liquid pair has this behaviour.

A nonvolatile solute contributes negligibly to the vapour pressure. For an ideal solution with a volatile solvent, only the solvent term remains. Its mole fraction is below one, so its vapour pressure is below the pure-solvent value. This is the starting point for relative lowering of vapour pressure.

1. Use the same pure pressures with $x_A=0.750$. Find total pressure.
2. What happens to solvent vapour pressure after adding a small amount of an ideal nonvolatile solute?

Answers

1. $67.5\,\mathrm{kPa}$.
2. It falls because the solvent mole fraction falls.

An ideal solution is a reference model. A real mixture can lie above or below that model, and the direction tells you something about molecular interactions.

An ideal solution obeys Raoult’s law over the full composition range at the stated temperature. Mixing does not change enthalpy or volume in the ideal model. Interactions between unlike molecules are similar enough to those between like molecules that neither component gains a strong preference for staying in the liquid.

$$\Delta H_{\rm mix}=0,\qquad\Delta V_{\rm mix}=0$$

For positive deviation, the observed partial pressures are higher than the Raoult reference. Molecules escape more readily than the ideal model predicts. A helpful molecular picture is that unlike interactions are weaker than the interactions they replace. Mixing is often endothermic and accompanied by expansion, but the measured pressure deviation is the defining observation.

For negative deviation, the observed pressures are lower than the reference. Stronger unlike interactions help keep molecules in the liquid. Mixing is often exothermic and accompanied by contraction. These descriptions compare a real mixture with a reference at the same temperature and composition.

For a model mixture, Raoult’s law predicts $40\,\mathrm{kPa}$, but the measured total pressure is $46\,\mathrm{kPa}$. Classify the deviation.

Compare observed and reference pressures directly. The observed value is higher by a positive amount.

Result: Positive deviation; the pressure difference is $6\,\mathrm{kPa}$.

An azeotrope is a mixture that boils with the same composition in liquid and vapour at a specified pressure. Ordinary fractional distillation cannot cross that azeotropic composition. A sufficiently strong positive deviation can produce a minimum-boiling azeotrope; a sufficiently strong negative deviation can produce a maximum-boiling one. Not every deviation is large enough to create an azeotrope.

1. A mixture has observed pressure below the Raoult reference. What is the sign of its deviation?
2. Can ordinary fractional distillation separate an azeotrope beyond its azeotropic composition?

Answers

1. Negative.
2. No. The vapour and liquid have the same composition at that point.

Some solution effects count particles rather than identify them. These colligative properties connect measurable changes to the amount of dissolved material.

Use the dilute, ideal-solution approximation unless a problem says otherwise. For an electrolyte, particle formation must also be considered. Begin with a nonelectrolyte that neither associates nor dissociates.

$$\frac{p^*-p}{p^*}=x_{\rm solute}\approx\frac{n_{\rm solute}}{n_{\rm solvent}}$$

The exact mole-fraction expression includes both amounts in the denominator. Replacing it with the solute-to-solvent mole ratio is a dilute approximation. A measured lowering can therefore give a solute amount and, from its mass, its molar mass.

Boiling occurs when the liquid vapour pressure reaches the external pressure. A nonvolatile solute lowers the solvent vapour pressure, so a higher temperature is needed to reach that pressure.

$$\Delta T_b=K_bm$$

Dissolved particles favour the liquid relative to the pure solid solvent. The equilibrium freezing temperature decreases. The constant depends on the solvent; molality supplies the solute amount per solvent mass.

$$\Delta T_f=K_fm$$

A nonelectrolyte of molar mass $100\,\mathrm{g\,mol^{-1}}$ has mass $10.0\,\mathrm g$ in $0.500\,\mathrm{kg}$ of water. Use $K_b=0.512$ and $K_f=1.86\,\mathrm{K\,kg\,mol^{-1}}$.

First find the molality. Use the same molality in both changes; the solvent constants differ.

Result: $m=0.2\,\mathrm{mol\,kg^{-1}}$, $\Delta T_b=0.1024\,\mathrm K$, and $\Delta T_f=0.372\,\mathrm K$.

A semipermeable membrane allows solvent to pass more readily than solute. If the two sides have different solvent chemical potentials, solvent flows towards the side with lower solvent chemical potential. In a simple dilute comparison at equal pressure, this is towards the more concentrated solution.

$$\pi=cRT=\frac{nRT}{V}$$

Osmotic pressure is the additional pressure needed to prevent net osmosis. Isotonic solutions have equal osmotic pressure at the same temperature. Hypertonic and hypotonic compare higher and lower osmotic pressure relative to another solution. Use absolute temperature and a gas constant whose units match pressure and volume.

A macromolecule sample of mass $0.120\,\mathrm g$ forms $0.250\,\mathrm L$ of solution at $300\,\mathrm K$. Its osmotic pressure is $0.00300\,\mathrm{bar}$. Use $R=0.08314\,\mathrm{L\,bar\,mol^{-1}\,K^{-1}}$.

Rearrange $\pi=wRT/(MV)$ to isolate the molar mass. Osmotic pressure is useful for large molecules because it can be measurable without a large temperature change.

Result: $M=3990.72\,\mathrm{g\,mol^{-1}}$, about $3.99\times10^3\,\mathrm{g\,mol^{-1}}$.

Apply pressure greater than the relevant osmotic-pressure difference to the concentrated side. Solvent can then move in the direction opposite to ordinary osmosis. A suitable membrane retains many dissolved substances. Membrane selectivity matters; the statement does not mean every contaminant is removed equally well.

1. Find osmotic pressure for $c=0.0100\,\mathrm{mol\,L^{-1}}$ at $300\,\mathrm K$ with the gas constant above.
2. What pressure direction is needed for reverse osmosis?

Answers

1. $0.24942\,\mathrm{bar}$.
2. An extra pressure on the concentrated side, large enough to exceed the osmotic-pressure difference.

If a solute splits into several particles, a particle-counting measurement can make its apparent molar mass look too small. Association produces the opposite effect.

The van ’t Hoff factor compares the observed colligative effect with the effect predicted if the solute stayed as single formula units. It is also the ratio of the normal molar mass to the apparent molar mass inferred using that assumption.

$$i=\frac{\text{observed effect}}{\text{effect without particle change}}=\frac{M_{\rm normal}}{M_{\rm apparent}}$$

$$\Delta T_f=iK_fm,\qquad\Delta T_b=iK_bm,\qquad\pi=icRT$$

Dissociation often gives a factor greater than one. Association often gives a factor below one. Real ionic solutions can depart from simple integer particle counts because ions interact. State the model and concentration range when interpreting a measured factor.

If one formula unit dissociates into $\nu$ particles and a fraction $\alpha$ dissociates, begin with one mole of formula units. Undissociated units contribute $1-\alpha$ moles and ions contribute $\nu\alpha$ moles.

$$i=1+\alpha(\nu-1)$$

A dilute salt that forms two ions per formula unit has measured $i=1.85$. Interpret this using the simple dissociation model.

Use $\nu=2$ and isolate $\alpha$. Treat the result as a model estimate rather than proof that every reduction in particle effect is incomplete dissociation.

Result: $\alpha=0.85$, or $85\%$.

For association of single molecules into dimers, each associated pair becomes one particle. If $\alpha$ is the fraction of original molecules that associates, $i=1-\alpha/2$. The factor is not zero even when every molecule is paired.

A solute that can form dimers has $i=0.800$. Find its associated fraction in the dimer-only model.

Rearrange $i=1-\alpha/2$ to $\alpha=2(1-i)$.

Result: $\alpha=0.4$, or $40\%$.

1. A solute has normal molar mass $150$ and apparent molar mass $300\,\mathrm{g\,mol^{-1}}$. Find $i$.
2. Why does association make the apparent molar mass larger?

Answers

1. $i=0.5$; association is consistent with this result.
2. The same mass produces fewer particles, so a particle-counting calculation infers fewer moles and therefore a larger molar mass.

Keep $x_A+x_B=1$, $p_{\rm total}=p_A+p_B$ and the direction of each temperature change visible in your working.

Try these original questions without looking at the answers. Show equations, conditions and units when relevant.

1.1 Give a homogeneous gas-mixture example and explain why a gas mixture can be a solution.

1.2 Distinguish solute identity from the role “solute”.

1.3 Find the mass percentage when $12\,\mathrm g$ of solute is mixed with $188\,\mathrm g$ of solvent.

1.4 A final solution volume of $0.600\,\mathrm L$ contains $0.180\,\mathrm{mol}$ of solute. Find molarity.

1.5 Find molality for $0.120\,\mathrm{mol}$ in $0.300\,\mathrm{kg}$ of solvent.

1.6 Find the solute mole fraction for $0.200\,\mathrm{mol}$ of solute and $4.80\,\mathrm{mol}$ of solvent.

1.7 Convert a mass fraction of $0.0000080$ to ppm by mass.

1.8 Explain why molarity can change when a closed solution is warmed.

1.9 What final volume is needed to dilute $60\,\mathrm{mL}$ of $0.400\,\mathrm M$ solution to $0.150\,\mathrm M$?

1.10 Does a higher stirring speed always increase equilibrium solubility? Explain.

1.11 Predict the usual temperature response for an endothermic dissolution.

1.12 Use $K_H=1.6\times10^5\,\mathrm{bar}$ and $p=8.0\,\mathrm{bar}$ to find dissolved mole fraction.

1.13 Why must a Henry-law question specify its convention?

1.14 A gas dissolves and reacts strongly with the solvent. Is a simple Henry-law calculation automatically reliable?

1.15 For an ideal binary solution, $x_A=0.200$, $p_A^*=90$ and $p_B^*=40\,\mathrm{kPa}$. Find the partial pressures.

1.16 Find total pressure and vapour fraction of A for the preceding mixture.

1.17 Can liquid and vapour have different composition at equilibrium?

1.18 A solvent has $p^*=24.0\,\mathrm{kPa}$ and mole fraction $0.950$. Find its ideal-solution vapour pressure.

1.19 State the reference used to identify positive deviation.

1.20 A mixture has stronger unlike attractions than expected. Which deviation is commonly consistent with this?

1.21 Why does an azeotrope limit ordinary fractional distillation?

1.22 A pure solvent has pressure $25.0\,\mathrm{kPa}$ and its solution has pressure $24.5\,\mathrm{kPa}$. Find relative lowering.

1.23 For the preceding ideal nonvolatile-solute mixture, what is solute mole fraction?

1.24 Find boiling-point elevation for $m=0.350\,\mathrm{mol\,kg^{-1}}$ and $K_b=0.512\,\mathrm{K\,kg\,mol^{-1}}$.

1.25 Find freezing-point depression for $m=0.350\,\mathrm{mol\,kg^{-1}}$ and $K_f=1.86\,\mathrm{K\,kg\,mol^{-1}}$.

1.26 A solvent freezes at $273.15\,\mathrm K$. Its solution has a depression of $0.744\,\mathrm K$. Find the solution freezing point.

1.27 A solute of mass $3.00\,\mathrm g$ in $0.200\,\mathrm{kg}$ of water gives $\Delta T_f=0.310\,\mathrm K$. Use $K_f=1.86$ and $i=1$. Find molar mass.

1.28 Why is osmotic pressure often useful for macromolecules?

1.29 Find osmotic pressure of $0.0150\,\mathrm M$ nonelectrolyte at $298\,\mathrm K$, using $R=0.08314\,\mathrm{L\,bar\,mol^{-1}\,K^{-1}}$.

1.30 Define isotonic at a fixed temperature.

1.31 Describe the direction of solvent movement in ordinary osmosis for two dilute solutions at equal pressure.

1.32 What reverses that direction in reverse osmosis?

1.33 An electrolyte has observed freezing-point depression $0.744\,\mathrm K$ when the undissociated prediction is $0.400\,\mathrm K$. Find $i$.

1.34 For a formula unit forming three ions, find ideal-model $i$ when $\alpha=0.650$.

1.35 For a three-ion dissociation, find $\alpha$ when $i=2.40$.

1.36 For dimer formation, find $i$ when half of the original molecules associate.

1.37 Normal molar mass is $200$ and apparent molar mass is $125\,\mathrm{g\,mol^{-1}}$. Find $i$.

1.38 Why can the observed factor of an ionic solute be below its ideal ion count?

1.39 Which solvent mass belongs in a molality calculation after two solutions are mixed?

1.40 A solution has the expected concentration but a lower-than-expected osmotic pressure. Give two model issues to examine.

1.41 Design a sequence for finding molar mass from a colligative measurement.

Answers

1.1 A uniform helium–oxygen mixture is homogeneous at molecular scale; both components are gases.

1.2 A substance is called a solute because of its role in a particular mixture. The same substance may be solvent in another mixture.

1.3 $6\%$.

1.4 $0.3\,\mathrm{mol\,L^{-1}}$.

1.5 $0.4\,\mathrm{mol\,kg^{-1}}$.

1.6 $0.04$.

1.7 $8\,\mathrm{ppm}$.

1.8 The solution volume can expand while its solute amount stays fixed.

1.9 $160\,\mathrm{mL}$.

1.10 No. It often speeds approach to equilibrium without changing the equilibrium composition.

1.11 Solubility commonly increases with temperature; the conclusion uses the stated positive dissolution enthalpy.

1.12 $0.00005$.

1.13 Some conventions put concentration or mole fraction on the other side, so the interpretation of a larger constant changes.

1.14 No. A reaction can change the dissolved species and invalidate the simple nonreactive dilute model.

1.15 $p_A=18$ and $p_B=32\,\mathrm{kPa}$.

1.16 $p=50\,\mathrm{kPa}$ and $y_A=0.36$.

1.17 Yes. Equality of phase equilibrium conditions does not require equal mole fractions in both phases.

1.18 $22.8\,\mathrm{kPa}$.

1.19 Compare measured partial pressures with Raoult-law predictions at the same temperature and liquid composition.

1.20 Negative deviation: molecules escape less readily, so vapour pressure is below the ideal reference.

1.21 At its azeotropic composition, equilibrium vapour has the same composition as the boiling liquid.

1.22 $0.02$.

1.23 It equals the relative lowering, so it has the same value as the previous answer.

1.24 $0.1792\,\mathrm K$ for a nonelectrolyte.

1.25 $0.651\,\mathrm K$ for a nonelectrolyte.

1.26 $272.406\,\mathrm K$.

1.27 $M=90\,\mathrm{g\,mol^{-1}}$.

1.28 It can give a measurable effect at low concentration and moderate temperature, where boiling/freezing changes may be too small.

1.29 $0.37164\,\mathrm{bar}$.

1.30 Two solutions are isotonic relative to each other when they have equal osmotic pressures.

1.31 Across a suitable membrane, solvent moves towards the side of higher solute particle concentration.

1.32 Sufficient extra pressure applied to the concentrated side overcomes the osmotic-pressure difference.

1.33 $1.86$.

1.34 $i=2.3$.

1.35 $\alpha=0.7$.

1.36 $i=0.75$.

1.37 $i=1.6$, consistent with an increased particle count.

1.38 Ionic interactions and nonideality make particle effects differ from the independent-ion limiting model.

1.39 Use the total mass of solvent, not total solution mass, while adding the solute amounts consistently.

1.40 Check possible association and nonideality, as well as temperature, membrane behaviour and measurement units.

1.41 Identify the solvent constant or gas constant, convert units, find particle amount from the measured effect, account for any stated particle factor, then divide solute mass by solute amount.

Quick-check answers appear directly after each topic. The selected explanations below show how to check the reasoning, rather than memorise a final line.

Normal molar mass is $200$ and apparent molar mass is $125\,\mathrm{g\,mol^{-1}}$. Find $i$.

$i=1.6$, consistent with an increased particle count.

Why can the observed factor of an ionic solute be below its ideal ion count?

Ionic interactions and nonideality make particle effects differ from the independent-ion limiting model.

Which solvent mass belongs in a molality calculation after two solutions are mixed?

Use the total mass of solvent, not total solution mass, while adding the solute amounts consistently.

A solution has the expected concentration but a lower-than-expected osmotic pressure. Give two model issues to examine.

Check possible association and nonideality, as well as temperature, membrane behaviour and measurement units.

Design a sequence for finding molar mass from a colligative measurement.

Identify the solvent constant or gas constant, convert units, find particle amount from the measured effect, account for any stated particle factor, then divide solute mass by solute amount.