Chapter 1: Solutions (Physical Chemistry)

Mass of solution $=22+122=144$ g.

• Benzene: $\frac{22}{144}\times 100=15.28\%$
• $\mathrm{CCl}{}_4$: $\frac{122}{144}\times 100=84.72\%$

Benzene 15.28%, CCl₄ 84.72%.

Take 100 g of solution: 30 g benzene and 70 g $\mathrm{CCl}{}_4$.

• $n(\mathrm{C}{}_6\mathrm{H}{}_6)=\frac{30}{78}=0.385$ mol
• $n(\mathrm{CCl}{}_4)=\frac{70}{154}=0.455$ mol

$$x_{\text{benzene}}=\frac{0.385}{0.385+0.455}=\frac{0.385}{0.840}\approx 0.459$$

$x_{\text{benzene}}\approx 0.459$ (and $x_{{\text{CCl}}_4}\approx 0.541$).

(a) Molar mass of $\mathrm{Co}(\mathrm{NO}{}_3){}_2\cdot 6\mathrm{H}{}_2\mathrm{O}=58.9+2(62)+6(18)\approx 291$ g mol⁻¹.

$n=\frac{30}{291}=0.103$ mol; $M=\frac{0.103}{4.3}\approx 0.024$ M.

(b) Moles of acid do not change on dilution: $M_1{V}_1=M_2{V}_2$:

$0.5\times 30=M_2\times 500$, so $M_2=0.03$ M.

(a) 0.024 M. (b) 0.03 M.

0.25 molal means 0.25 mol of urea (0.25 × 60 = 15 g) in 1000 g of water, i.e. in 1015 g of solution.

$$\text{urea in 2500 g of solution}=\frac{15}{1015}\times 2500\approx 36.95\text{g}$$

About 36.95 g.

Take 100 g of solution: 20 g KI ($\frac{20}{166}=0.1205$ mol) and 80 g water ($\frac{80}{18}=4.444$ mol).

(a) $m=\frac{0.1205}{0.080}\approx 1.51$ mol kg⁻¹.

(b) Volume of 100 g of solution $=\frac{100}{1.202}=83.2$ mL, so $M=\frac{0.1205}{0.0832}\approx 1.45$ mol L⁻¹.

(c) $x_{\text{KI}}=\frac{0.1205}{0.1205+4.444}\approx 0.0264$.

(a) 1.51 m. (b) 1.45 M. (c) 0.0264.

0.195 mol of $\mathrm{H}{}_2\mathrm{S}$ in 1 kg (55.56 mol) of water:

$$x=\frac{0.195}{0.195+55.56}\approx 3.50\times 10^{-3}$$

At STP, $p=0.987$ bar. $K_H=\frac{p}{x}=\frac{0.987}{3.50\times 10^{-3}}\approx 282$ bar.

$K_H\approx 282$ bar.

$p=2.5\times 1.01325\times 10^5=2.533\times 10^5$ Pa.

$x=\frac{p}{K_H}=\frac{2.533\times 10^5}{1.67\times 10^8}=1.517\times 10^{-3}$.

500 mL of water ≈ 500 g ≈ 27.78 mol. For such a small $x$, $n(\mathrm{CO}{}_2)\approx x\times n(\text{water})=1.517\times 10^{-3}\times 27.78=0.0421$ mol.

Mass $=0.0421\times 44\approx 1.85$ g.

About 1.85 g of CO₂.

Liquid: $p=p_A^{\circ }{x}_A+p_B^{\circ }(1-x_A)$:

$600=450x_A+700(1-x_A)$, so $250x_A=100$, $x_A=0.40$, $x_B=0.60$.

Vapour: $p_A=450\times 0.40=180$ mm, $p_B=700\times 0.60=420$ mm.

$y_A=\frac{180}{600}=0.30$, $y_B=\frac{420}{600}=0.70$.

Liquid: $x_A$ = 0.40, $x_B$ = 0.60. Vapour: $y_A$ = 0.30, $y_B$ = 0.70.

$n(\text{urea})=\frac{50}{60}=0.833$ mol; $n(\text{water})=\frac{850}{18}=47.22$ mol.

Relative lowering $=x_{\text{urea}}=\frac{0.833}{0.833+47.22}\approx 0.0173$.

$p=p^{\circ }(1-0.0173)=23.8\times 0.9827\approx 23.4$ mm Hg.

p ≈ 23.4 mm Hg; relative lowering ≈ 0.0173.

$\Delta T_b=100-99.63=0.37$ K; $K_b(\text{water})=0.52$ K kg mol⁻¹.

$m=\frac{\Delta T_b}{K_b}=\frac{0.37}{0.52}=0.712$ mol kg⁻¹. For 500 g of water: $0.356$ mol.

Mass of sucrose $=0.356\times 342\approx 121.7$ g.

About 121.7 g.

$m=\frac{\Delta T_f}{K_f}=\frac{1.5}{3.9}=0.385$ mol kg⁻¹. In 75 g: $0.385\times 0.075=0.0288$ mol.

Molar mass of $\mathrm{C}{}_6\mathrm{H}{}_8\mathrm{O}{}_6=176$ g mol⁻¹: mass $=0.0288\times 176\approx 5.08$ g.

About 5.08 g.

$n=\frac{1.0}{185000}=5.41\times 10^{-6}$ mol; $V=450$ mL $=4.5\times 10^{-4}$ m³; $T=310$ K.

$$\Pi =\frac{nRT}{V}=\frac{5.41\times 10^{-6}\times 8.314\times 310}{4.5\times 10^{-4}}\approx 31\text{Pa}$$

About 31 Pa (30.96 Pa).

A solution is a homogeneous mixture of two or more substances whose composition can vary within limits. The component present in the larger amount is the solvent; the others are solutes.

A solute and a solvent can each be a gas, a liquid or a solid, so there are nine types of solutions:

A homogeneous mixture of two or more substances; nine types (gaseous, liquid and solid solutions, each with a gas, liquid or solid solute), as in the table.

Hydrogen dissolved in palladium (palladium absorbs large volumes of hydrogen).

Hydrogen in palladium.

• (i) Mole fraction of a component: moles of that component divided by the total moles of all components. $x_A=\frac{n_A}{n_A+n_B+\dots }$ (no unit; all mole fractions add to 1).
• (ii) Molality (m): moles of solute per kilogram of solvent. Unit mol kg⁻¹; it does not change with temperature.
• (iii) Molarity (M): moles of solute per litre of solution. Unit mol L⁻¹; it changes slightly with temperature (volume changes).
• (iv) Mass percentage: mass of a component per 100 g of solution: $\frac{\text{mass of component}}{\text{total mass of solution}}\times 100$.

As defined above.

In 100 g of solution: 68 g $\mathrm{HNO}{}_3$ $=\frac{68}{63}=1.079$ mol.

Volume of 100 g of solution $=\frac{100}{1.504}=66.5$ mL $=0.0665$ L.

$M=\frac{1.079}{0.0665}\approx 16.2$ M.

About 16.2 M.

In 100 g: 10 g glucose $=\frac{10}{180}=0.0556$ mol; 90 g water $=5.0$ mol.

• Molality $=\frac{0.0556}{0.090}\approx 0.617$ mol kg⁻¹
• $x_{\text{glucose}}=\frac{0.0556}{5.0556}\approx 0.011$, $x_{\text{water}}\approx 0.989$
• Volume $=\frac{100}{1.2}=83.3$ mL, molarity $=\frac{0.0556}{0.0833}\approx 0.67$ M

m ≈ 0.617 mol kg⁻¹; x(glucose) ≈ 0.011, x(water) ≈ 0.989; M ≈ 0.67 mol L⁻¹.

Let each be $x$ mol: $106x+84x=1$, so $x=\frac{1}{190}=5.26\times 10^{-3}$ mol.

• $\mathrm{Na}{}_2\mathrm{CO}{}_3+2\mathrm{HCl}\to 2\mathrm{NaCl}+\mathrm{H}{}_2\mathrm{O}+\mathrm{CO}{}_2$: needs $2x$ mol HCl
• $\mathrm{NaHCO}{}_3+\mathrm{HCl}\to \mathrm{NaCl}+\mathrm{H}{}_2\mathrm{O}+\mathrm{CO}{}_2$: needs $x$ mol HCl

Total HCl $=3x=0.0158$ mol. Volume $=\frac{0.0158}{0.1}=0.158$ L $=158$ mL.

About 158 mL.

Solute: $300\times 0.25+400\times 0.40=75+160=235$ g in $700$ g of solution.

Mass % of solute $=\frac{235}{700}\times 100\approx 33.6\%$ (water 66.4%).

Solute about 33.6%, water about 66.4%.

$n=\frac{222.6}{62}=3.59$ mol.

Molality $=\frac{3.59}{0.200}\approx 17.95$ mol kg⁻¹.

Mass of solution $=422.6$ g; volume $=\frac{422.6}{1.072}=394.2$ mL. Molarity $=\frac{3.59}{0.3942}\approx 9.11$ M.

Molality ≈ 17.95 m; molarity ≈ 9.11 M.

15 ppm = 15 g of chloroform in $10^6$ g of solution.

(i) $\frac{15}{10^6}\times 100=1.5\times 10^{-3}\%$.

(ii) $n=\frac{15}{119.5}=0.1255$ mol in about $10^6$ g $=1000$ kg of water:

molality $=\frac{0.1255}{1000}\approx 1.25\times 10^{-4}$ mol kg⁻¹.

(i) $1.5\times 10^{-3}$%. (ii) About $1.25\times 10^{-4}$ m.

Both alcohol and water are hydrogen-bonded liquids. When they mix, the new alcohol–water hydrogen bonds are weaker than the original water–water and alcohol–alcohol interactions. So molecules escape more easily, the vapour pressure is higher than Raoult's law predicts, and the solution shows positive deviation. (The mixing is slightly endothermic, and ethanol and water form a minimum-boiling azeotrope at about 95% ethanol.)

Alcohol–water interactions are weaker than in the pure liquids, so the solution shows positive deviation from Raoult's law.

Dissolving a gas in a liquid is an exothermic process (gas + liquid ⇌ solution + heat). By Le Chatelier's principle, raising the temperature shifts this equilibrium backwards, so less gas stays dissolved. (Also, the gas molecules have more kinetic energy and escape more easily.)

Dissolution of a gas is exothermic, so heating shifts the equilibrium back and solubility falls.

Henry's law: at constant temperature, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid. In terms of mole fraction: $p=K_{H}x$.

Applications:

• Soft drinks and soda water are bottled under high $\mathrm{CO}{}_2$ pressure to dissolve more $\mathrm{CO}{}_2$.
• Scuba divers breathe air diluted with helium to avoid "the bends" (nitrogen dissolved at high pressure forms bubbles in the blood when they surface).
• At high altitudes the low partial pressure of oxygen means less oxygen in the blood, causing weakness (anoxia).

Solubility of a gas ∝ its partial pressure ($p=K_{H}x$); used in bottling soft drinks, deep-sea diving gas mixtures, and explains anoxia at high altitude.

By Henry's law, the amount dissolved is proportional to the partial pressure:

$$p_2=1\text{bar}\times \frac{5.00\times 10^{-2}}{6.56\times 10^{-3}}\approx 7.62\text{bar}$$

About 7.62 bar.

• Positive deviation: the vapour pressure of the solution is higher than Raoult's law predicts. The A–B attractions are weaker than A–A and B–B attractions, so molecules escape more easily. Mixing absorbs heat: ${\Delta }_{\text{mix}}H>0$ (and ${\Delta }_{\text{mix}}V>0$). Example: ethanol + acetone.
• Negative deviation: the vapour pressure is lower than predicted. The A–B attractions are stronger than A–A and B–B, so molecules escape less easily. Mixing releases heat: ${\Delta }_{\text{mix}}H<0$ (and ${\Delta }_{\text{mix}}V<0$). Example: chloroform + acetone.

Positive deviation: higher vapour pressure, weaker A–B forces, ${\Delta }_{\text{mix}}H>0$. Negative deviation: lower vapour pressure, stronger A–B forces, ${\Delta }_{\text{mix}}H<0$.

At the normal boiling point of water, $p^{\circ }=1.013$ bar. In 100 g of solution: 2 g solute and 98 g water.

$$\frac{p^{\circ }-p}{p^{\circ }}=\frac{n_2}{n_1}\Rightarrow \frac{1.013-1.004}{1.013}=\frac{2/M_2}{98/18}$$

$0.00888=\frac{2\times 18}{98M_2}$, so $M_2=\frac{36}{98\times 0.00888}\approx 41.35$ g mol⁻¹.

About 41.35 g mol⁻¹.

$n(\text{heptane})=\frac{26.0}{100}=0.260$ mol; $n(\text{octane})=\frac{35}{114}=0.307$ mol.

$x_{\text{heptane}}=\frac{0.260}{0.567}=0.459$; $x_{\text{octane}}=0.541$.

$$p=105.2\times 0.459+46.8\times 0.541=48.3+25.3\approx 73.6\text{kPa}$$

About 73.6 kPa.

1 molal: 1 mol of solute in 1000 g (55.56 mol) of water.

$x_{\text{solute}}=\frac{1}{56.56}=0.0177$, so $p=12.3\times (1-0.0177)\approx 12.08$ kPa.

About 12.08 kPa.

$p=0.80p^{\circ }$, so the relative lowering is $0.20$. $n(\text{octane})=\frac{114}{114}=1$ mol.

Using the textbook's dilute-solution form $\frac{p^{\circ }-p}{p^{\circ }}=\frac{n_2}{n_1}$:

$0.20=\frac{w/40}{1}$, so $w=8$ g.

(If the exact form $x_2=\frac{n_2}{n_1+n_2}=0.20$ is used instead, $n_2=0.25$ mol and $w=10$ g. The NCERT answer uses the first form: 8 g.)

8 g (by the textbook method; 10 g by the exact mole-fraction form).

Let the molar mass be $M$, so $n_2=\frac{30}{M}$. Water: 90 g = 5 mol, then 108 g = 6 mol. Using $\frac{p^{\circ }-p}{p^{\circ }}=\frac{n_2}{n_1}$:

• First solution: $1-\frac{2.8}{p^{\circ }}=\frac{n_2}{5}$, so $n_2=5-\frac{14}{p^{\circ }}$
• After adding water: $1-\frac{2.9}{p^{\circ }}=\frac{n_2}{6}$, so $n_2=6-\frac{17.4}{p^{\circ }}$

Equating: $5-\frac{14}{p^{\circ }}=6-\frac{17.4}{p^{\circ }}$, so $\frac{3.4}{p^{\circ }}=1$ and $p^{\circ }=3.4$ kPa.

Then $n_2=5-\frac{14}{3.4}=0.882$ mol and $M=\frac{30}{0.882}\approx 34$ g mol⁻¹.

(i) About 34 g mol⁻¹. (ii) 3.4 kPa.

Both solutions have 5 g of solute in 95 g of water, so $\Delta T_f\propto \frac{1}{M}$.

$\Delta T_f(\text{sugar})=273.15-271=2.15$ K.

$$\Delta T_f(\text{glucose})=2.15\times \frac{342}{180}\approx 4.09\text{K}$$

Freezing point $=273.15-4.09\approx 269.06$ K.

About 269.06 K.

$M=\frac{K_f{w}_2\times 1000}{\Delta T_f{w}_1}$:

• $M(\mathrm{AB}{}_2)=\frac{5.1\times 1\times 1000}{2.3\times 20}\approx 110.9$ g mol⁻¹
• $M(\mathrm{AB}{}_4)=\frac{5.1\times 1\times 1000}{1.3\times 20}\approx 196.2$ g mol⁻¹

So $A+2B=110.9$ and $A+4B=196.2$. Subtracting: $2B=85.3$, $B\approx 42.6$; then $A\approx 25.6$.

A ≈ 25.6 u, B ≈ 42.6 u.

$\Pi =CRT$, so at the same temperature $\Pi \propto C$. $C_1=\frac{36}{180}=0.20$ M.

$$C_2=0.20\times \frac{1.52}{4.98}\approx 0.061\text{M}$$

About 0.061 M.

• (i) London dispersion forces (both non-polar)
• (ii) London dispersion forces (both non-polar)
• (iii) Ion–dipole interactions
• (iv) Dipole–dipole interactions (with some hydrogen bonding between the O–H of methanol and the C=O of acetone)
• (v) Dipole–dipole interactions

(i) dispersion; (ii) dispersion; (iii) ion–dipole; (iv) dipole–dipole; (v) dipole–dipole.

n-Octane is non-polar, so "like dissolves like": the more polar the solute, the less it dissolves.

KCl < $\mathrm{CH}{}_3\mathrm{OH}$ < $\mathrm{CH}{}_3\mathrm{CN}$ < cyclohexane

KCl is ionic (least soluble); methanol is polar and hydrogen-bonded; acetonitrile is polar but has no hydrogen bonding, so it is less held together; cyclohexane is non-polar like octane (most soluble).

KCl < CH₃OH < CH₃CN < cyclohexane.

• Insoluble: toluene, chloroform (non-polar or weakly polar, no hydrogen bonding with water)
• Partially soluble: phenol, pentanol (the –OH forms hydrogen bonds, but the large hydrocarbon part is non-polar)
• Highly soluble: formic acid, ethylene glycol (small molecules that form strong hydrogen bonds with water)

Insoluble: toluene, chloroform. Partially soluble: phenol, pentanol. Highly soluble: formic acid, ethylene glycol.

$n(\mathrm{Na}{}^{+})=\frac{92}{23}=4$ mol per kg.

Taking 1 kg of the lake water, its volume is $\frac{1000}{1.25}=800$ mL $=0.8$ L, so

$$M=\frac{4}{0.8}=5\text{M}$$

About 5 M.

$\mathrm{CuS}\rightleftharpoons \mathrm{Cu}{}^{2+}+\mathrm{S}{}^{2-}$, so $K_{sp}=s^2$:

$$s=\sqrt{6\times 10^{-16}}\approx 2.45\times 10^{-8}\text{M}$$

About $2.45\times 10^{-8}$ M.

$$\text{Mass \%}=\frac{6.5}{6.5+450}\times 100=\frac{6.5}{456.5}\times 100\approx 1.42\%$$

About 1.42%.

Molar mass of $\mathrm{C}{}_{19}\mathrm{H}{}_{21}\mathrm{NO}{}_3=19(12)+21(1)+14+3(16)=311$ g mol⁻¹.

$1.5\times 10^{-3}$ m means $1.5\times 10^{-3}\times 311=0.4665$ g of nalorphene per 1000 g of water.

For 1.5 mg $=1.5\times 10^{-3}$ g: water needed $=\frac{1.5\times 10^{-3}}{0.4665}\times 1000=3.215$ g.

Mass of solution $\approx 3.215+0.0015\approx 3.22$ g.

About 3.22 g of solution.

$n=0.15\times 0.250=0.0375$ mol; molar mass of $\mathrm{C}{}_6\mathrm{H}{}_5\mathrm{COOH}=122$ g mol⁻¹.

Mass $=0.0375\times 122\approx 4.58$ g.

About 4.58 g.

$\Delta T_f$ depends on the number of particles, so it is larger for an acid that ionises more. Cl and F are electron-withdrawing; they pull electron density away from the –COOH group and make the $\mathrm{H}{}^{+}$ easier to lose. F is more electronegative than Cl, so:

acid strength (and ionisation): $\mathrm{CH}{}_3\mathrm{COOH}<\mathrm{CCl}{}_3\mathrm{COOH}<\mathrm{CF}{}_3\mathrm{COOH}$

More ionisation means more particles (a larger van't Hoff factor $i$), so the depression of freezing point increases in the same order.

The acid strength (degree of ionisation) increases in that order because of the electron-withdrawing Cl and F, giving more particles and a larger ΔT_f.

Molar mass of $\mathrm{CH}{}_3\mathrm{CH}{}_2\mathrm{CHClCOOH}=122.5$ g mol⁻¹. $n=\frac{10}{122.5}=0.0816$ mol; $m=\frac{0.0816}{0.250}=0.3265$ mol kg⁻¹.

Degree of dissociation from $K_a=\frac{c{\alpha }^2}{1-\alpha }$:

$0.3265{\alpha }^2+1.4\times 10^{-3}\alpha -1.4\times 10^{-3}=0$, giving $\alpha \approx 0.0633$.

$i=1+\alpha =1.0633$.

$$\Delta T_f=i{K}_{f}m=1.0633\times 1.86\times 0.3265\approx 0.65\text{K}$$

About 0.65 K.

Molar mass of $\mathrm{CH}{}_2\mathrm{FCOOH}=78$ g mol⁻¹. $n=\frac{19.5}{78}=0.25$ mol; $m=\frac{0.25}{0.5}=0.5$ mol kg⁻¹.

Expected (no dissociation): $\Delta T_f=1.86\times 0.5=0.93$ K.

$$i=\frac{1.0}{0.93}\approx 1.075$$

For $\mathrm{HA}\rightleftharpoons \mathrm{H}{}^{+}+\mathrm{A}{}^{-}$, $i=1+\alpha$, so $\alpha =0.075$.

$$K_a=\frac{c{\alpha }^2}{1-\alpha }=\frac{0.5\times (0.075{)}^2}{0.925}\approx 3.07\times 10^{-3}$$

i ≈ 1.075; $K_a\approx 3.07\times 10^{-3}$.

$n(\text{glucose})=\frac{25}{180}=0.139$ mol; $n(\text{water})=\frac{450}{18}=25$ mol.

$x_{\text{glucose}}=\frac{0.139}{25.139}=0.00553$.

$p=17.535\times (1-0.00553)\approx 17.44$ mm Hg.

About 17.44 mm Hg.

$$x=\frac{p}{K_H}=\frac{760}{4.27\times 10^5}\approx 1.78\times 10^{-3}$$

Mole fraction of methane ≈ $1.78\times 10^{-3}$.

$n_A=\frac{100}{140}=0.714$ mol; $n_B=\frac{1000}{180}=5.556$ mol; $x_A=\frac{0.714}{6.270}=0.114$, $x_B=0.886$.

$p_B=500\times 0.886=443.0$ torr, so $p_A=475-443.0=32.0$ torr.

$p_A^{\circ }=\frac{p_A}{x_A}=\frac{32.0}{0.114}\approx 280.7$ torr.

$p_A^{\circ }$ ≈ 280.7 torr; $p_A$ in the solution ≈ 32 torr.

For an ideal solution: $p_{\text{acetone}}=741.8x$, $p_{\text{chloroform}}=632.8(1-x)$ and $p_{\text{total}}=632.8+109x$ (straight lines). Comparing totals:

On the graph, the experimental curves for $p_{\text{acetone}}$, $p_{\text{chloroform}}$ and $p_{\text{total}}$ all lie below the straight ideal lines (the total curve sags to a minimum near $x\approx 0.36$). So the solution shows negative deviation: acetone and chloroform attract each other through hydrogen bonding ($\mathrm{Cl}{}_3\mathrm{C}-\mathrm{H}\cdot \cdot \cdot \mathrm{O}=\mathrm{C}$).

The experimental vapour pressures are lower than the ideal values at every composition: negative deviation from Raoult's law.

$n(\text{benzene})=\frac{80}{78}=1.026$ mol; $n(\text{toluene})=\frac{100}{92}=1.087$ mol; $x_{\text{benzene}}=\frac{1.026}{2.113}=0.486$.

$p_{\text{benzene}}=50.71\times 0.486=24.6$ mm; $p_{\text{toluene}}=32.06\times 0.514=16.5$ mm; total $=41.1$ mm.

$$y_{\text{benzene}}=\frac{24.6}{41.1}\approx 0.60$$

About 0.60.

Total pressure $=10$ atm $=7600$ mm Hg.

• $p(\mathrm{O}{}_2)=0.20\times 7600=1520$ mm: $x(\mathrm{O}{}_2)=\frac{1520}{3.30\times 10^7}\approx 4.6\times 10^{-5}$
• $p(\mathrm{N}{}_2)=0.79\times 7600=6004$ mm: $x(\mathrm{N}{}_2)=\frac{6004}{6.51\times 10^7}\approx 9.2\times 10^{-5}$

x(O₂) ≈ $4.6\times 10^{-5}$; x(N₂) ≈ $9.2\times 10^{-5}$.

$\Pi =iCRT$, so

$$C=\frac{\Pi }{iRT}=\frac{0.75}{2.47\times 0.0821\times 300}\approx 0.0123{\text{mol L}}^{-1}$$

In 2.5 L: $n=0.0308$ mol. Molar mass of $\mathrm{CaCl}{}_2=111$ g mol⁻¹: mass $=0.0308\times 111\approx 3.42$ g.

About 3.42 g.

$\mathrm{K}{}_2\mathrm{SO}{}_4\to 2\mathrm{K}{}^{+}+\mathrm{SO}{}_4{}^{2-}$, so $i=3$. Molar mass $=174$ g mol⁻¹.

$n=\frac{0.025}{174}=1.44\times 10^{-4}$ mol; $C=\frac{1.44\times 10^{-4}}{2}=7.2\times 10^{-5}$ M; $T=298$ K.

$$\Pi =iCRT=3\times 7.2\times 10^{-5}\times 0.0821\times 298\approx 5.27\times 10^{-3}\text{atm}$$

About $5.27\times 10^{-3}$ atm.