The molality of a solution (denoted here by the symbol m) is a measure of concentration, defined as moles of solute divided by mass of solvent, or in other words m = (n_molute)/(Mn_solvent) where M is the solvent’s molar mass.

The molality of a solution (denoted here by the symbol m) is a measure of concentration, defined as moles of solute divided by mass of solvent, or in other words m = (n_molute)/(Mn_solvent) where M is the solvent’s molar mass. For aqueous solutions of NaCl at 25.0°C and 1 atm, and for 0 <= m <= 6 mol/(kg), the available volumetric data can be fit to the empirical function V = c0 + c1m + c2m^2 + c3m^3 with parameters c0 = 1001.70 cm^3 c1 = 17.298 kgcm^3/mol c2 = 0.9777 kg^2cm^3/mol^2 c3 = -0.0569 kg^3cm^3/mol^3. (a) Use this expression for V to derive an expression for the partial molar volume of NaCl. (b) Determine the partial molar volume of water (again in aqueous NaCl) as a function of m. (c) Given that the density of pure water (at 25°C and 1 atm) is 0.99707 g/cm^3, calculate the partial molar volume of water in a 1.250 molal solution of NaCl. What is the fractional change in hat(V)_water in this solution, relative to hat(V)_water? (d) Given 1.00 kg of a 1.250 molal solution of aqueous NaCl, suppose that an additional 10.0 g of NaCl is dissolved into this solution. Calculate the change in the volume of the solution. (e) What is the molar concentration (moles of NaCl per liter of solution) of a solution that is 1.250 molal sodium chloride?

The Correct Answer and Explanation is:

Let’s address each part step by step, clearly and concisely:

(a) Expression for the Partial Molar Volume of NaCl

We are given: V(m)=c0+c1m+c2m2+c3m3V(m) = c_0 + c_1m + c_2m^2 + c_3m^3V(m)=c0​+c1​m+c2​m2+c3​m3

where VVV is the volume of the solution per kg of solvent (in cm³), and mmm is the molality in mol/kg.

The partial molar volume of NaCl, V^NaCl\hat{V}_{\text{NaCl}}V^NaCl​, is given by: V^NaCl=(∂V∂m)T,P\hat{V}_{\text{NaCl}} = \left( \frac{\partial V}{\partial m} \right)_{T,P}V^NaCl​=(∂m∂V​)T,P​

Differentiate VVV with respect to mmm: V^NaCl=dVdm=c1+2c2m+3c3m2\hat{V}_{\text{NaCl}} = \frac{dV}{dm} = c_1 + 2c_2m + 3c_3m^2V^NaCl​=dmdV​=c1​+2c2​m+3c3​m2

(b) Partial Molar Volume of Water

Let VVV be total volume per 1 kg solvent. The total volume can be written as: V=nNaClV^NaCl+nH2OV^H2OV = n_{\text{NaCl}} \hat{V}_{\text{NaCl}} + n_{\text{H}_2O} \hat{V}_{\text{H}_2O}V=nNaCl​V^NaCl​+nH2​O​V^H2​O​

Given nNaCl=mn_{\text{NaCl}} = mnNaCl​=m and nH2O=1000/Mn_{\text{H}_2O} = 1000/MnH2​O​=1000/M where M=18.015 g/molM = 18.015 \text{ g/mol}M=18.015 g/mol, then: V^H2O=V−mV^NaClnH2O=V−mV^NaCl1000/18.015\hat{V}_{\text{H}_2O} = \frac{V – m\hat{V}_{\text{NaCl}}}{n_{\text{H}_2O}} = \frac{V – m\hat{V}_{\text{NaCl}}}{1000/18.015}V^H2​O​=nH2​O​V−mV^NaCl​​=1000/18.015V−mV^NaCl​​

(c) Partial Molar Volume of Water at 1.250 molal

First calculate VVV: V=1001.70+17.298(1.250)+0.9777(1.250)2−0.0569(1.250)3=1024.46 cm3V = 1001.70 + 17.298(1.250) + 0.9777(1.250)^2 – 0.0569(1.250)^3 = 1024.46 \text{ cm}^3V=1001.70+17.298(1.250)+0.9777(1.250)2−0.0569(1.250)3=1024.46 cm3

Now use: V^NaCl=c1+2c2m+3c3m2=17.298+2(0.9777)(1.25)+3(−0.0569)(1.25)2=19.56 cm3/mol\hat{V}_{\text{NaCl}} = c_1 + 2c_2m + 3c_3m^2 = 17.298 + 2(0.9777)(1.25) + 3(-0.0569)(1.25)^2 = 19.56 \text{ cm}^3/molV^NaCl​=c1​+2c2​m+3c3​m2=17.298+2(0.9777)(1.25)+3(−0.0569)(1.25)2=19.56 cm3/mol V^H2O=1024.46−1.250(19.56)1000/18.015=1024.46−24.4555.51≈18.01 cm3/mol\hat{V}_{\text{H}_2O} = \frac{1024.46 – 1.250(19.56)}{1000/18.015} = \frac{1024.46 – 24.45}{55.51} \approx 18.01 \text{ cm}^3/molV^H2​O​=1000/18.0151024.46−1.250(19.56)​=55.511024.46−24.45​≈18.01 cm3/mol

For pure water: V=1000 g0.99707 g/cm3=1002.94 cm3,V^H2O=1002.9455.51≈18.07 cm3/molV = \frac{1000 \text{ g}}{0.99707 \text{ g/cm}^3} = 1002.94 \text{ cm}^3, \quad \hat{V}_{\text{H}_2O} = \frac{1002.94}{55.51} \approx 18.07 \text{ cm}^3/molV=0.99707 g/cm31000 g​=1002.94 cm3,V^H2​O​=55.511002.94​≈18.07 cm3/mol

Fractional change: 18.01−18.0718.07≈−0.0033 or −0.33%\frac{18.01 – 18.07}{18.07} \approx -0.0033 \text{ or } -0.33\%18.0718.01−18.07​≈−0.0033 or −0.33%

(d) Volume Change After Adding 10.0 g NaCl to 1.250 molal Solution

Original moles NaCl = 1.250 mol
Add 10.0 g NaCl → 10.058.44≈0.1711\frac{10.0}{58.44} \approx 0.171158.4410.0​≈0.1711 mol
New total = 1.4211 mol
New molality = 1.4211 mol/kg

Use: ΔV=V(1.4211)−V(1.250)\Delta V = V(1.4211) – V(1.250)ΔV=V(1.4211)−V(1.250)

Compute: V(1.4211)=1001.70+17.298(1.4211)+0.9777(1.4211)2−0.0569(1.4211)3≈1027.23 cm3V(1.4211) = 1001.70 + 17.298(1.4211) + 0.9777(1.4211)^2 – 0.0569(1.4211)^3 \approx 1027.23 \text{ cm}^3V(1.4211)=1001.70+17.298(1.4211)+0.9777(1.4211)2−0.0569(1.4211)3≈1027.23 cm3 ΔV=1027.23−1024.46=2.77 cm3\Delta V = 1027.23 – 1024.46 = 2.77 \text{ cm}^3ΔV=1027.23−1024.46=2.77 cm3

(e) Molar Concentration of 1.250 molal NaCl

We know:

• 1.250 mol NaCl
• Volume from (c): 1024.46 cm³ = 1.02446 L

So: Molarity=1.250 mol1.02446 L≈1.22 mol/L\text{Molarity} = \frac{1.250 \text{ mol}}{1.02446 \text{ L}} \approx 1.22 \text{ mol/L}Molarity=1.02446 L1.250 mol​≈1.22 mol/L

Summary:

• V^NaCl=c1+2c2m+3c3m2\hat{V}_{\text{NaCl}} = c_1 + 2c_2m + 3c_3m^2V^NaCl​=c1​+2c2​m+3c3​m2
• V^H2O=V−mV^NaClnH2O\hat{V}_{\text{H}_2O} = \frac{V – m\hat{V}_{\text{NaCl}}}{n_{\text{H}_2O}}V^H2​O​=nH2​O​V−mV^NaCl​​
• V^H2O\hat{V}_{\text{H}_2O}V^H2​O​ decreases slightly (−0.33%) in 1.250 molal solution
• Adding 10 g NaCl increases volume by ~2.77 cm³
• Molarity of 1.250 molal NaCl ≈ 1.22 mol/L