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ChemMedMain 2026 • 4th Apr - Shift 1

Calculate Moles of Solvent from Elevation in Boiling Point and Vapour Pressure

A non-volatile, non-electrolyte solid solute when dissolved in 40 g40\text{ g} of a solvent, the vapour pressure of the solvent decreased from 760 mm Hg760\text{ mm Hg} to 750 mm Hg750\text{ mm Hg}. If the same solution boils at 320 K320\text{ K}, then the number of moles of the solvent present in the solution is _______. (Nearest integer) [Given: boiling point of the pure solvent =319.5 K= 319.5\text{ K}, Kb\text{K}_\text{b} of the solvent =0.3 K kg mol−1= 0.3\text{ K kg mol}^{-1}]

Official Numerical Answer5

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Step-by-Step Solution

To find the number of moles of the solvent present in the solution, we can use the concepts of elevation in boiling point and relative lowering of vapour pressure.

Step 1: Calculate the elevation in boiling point (ΔTb\Delta T_\text{b}) The elevation in boiling point is given by: ΔTb=Tb−Tb0\Delta T_\text{b} = T_\text{b} - T_\text{b}^0

Given:

  • Boiling point of the solution, Tb=320 KT_\text{b} = 320\text{ K}
  • Boiling point of the pure solvent, Tb0=319.5 KT_\text{b}^0 = 319.5\text{ K}

ΔTb=320 K−319.5 K=0.5 K\Delta T_\text{b} = 320\text{ K} - 319.5\text{ K} = 0.5\text{ K}

Step 2: Calculate the molality (mm) of the solution The relation between elevation in boiling point and molality is: ΔTb=Kb×m\Delta T_\text{b} = K_\text{b} \times m

Given Kb=0.3 K kg mol−1K_\text{b} = 0.3\text{ K kg mol}^{-1}: 0.5=0.3×m  ⟹  m=0.50.3=53 mol kg−10.5 = 0.3 \times m \implies m = \frac{0.5}{0.3} = \frac{5}{3}\text{ mol kg}^{-1}

Step 3: Calculate the number of moles of solute (nBn_\text{B}) Molality is defined as: m=nBWA (in kg)m = \frac{n_\text{B}}{W_\text{A}\text{ (in kg)}}

Given mass of solvent, WA=40 g=0.04 kgW_\text{A} = 40\text{ g} = 0.04\text{ kg}: 53=nB0.04  ⟹  nB=53×0.04=0.23=115 mol\frac{5}{3} = \frac{n_\text{B}}{0.04} \implies n_\text{B} = \frac{5}{3} \times 0.04 = \frac{0.2}{3} = \frac{1}{15}\text{ mol}

Step 4: Determine the number of moles of solvent (nAn_\text{A}) using Raoult's Law According to Raoult's Law for relative lowering of vapour pressure: P0−PP0=χB=nBnA+nB\frac{P^0 - P}{P^0} = \chi_\text{B} = \frac{n_\text{B}}{n_\text{A} + n_\text{B}}

Given:

  • Vapour pressure of pure solvent, P0=760 mm HgP^0 = 760\text{ mm Hg}
  • Vapour pressure of solution, P=750 mm HgP = 750\text{ mm Hg}

Substituting the values into the formula: 760−750760=nBnA+nB\frac{760 - 750}{760} = \frac{n_\text{B}}{n_\text{A} + n_\text{B}}

10760=nBnA+nB  ⟹  176=nBnA+nB\frac{10}{760} = \frac{n_\text{B}}{n_\text{A} + n_\text{B}} \implies \frac{1}{76} = \frac{n_\text{B}}{n_\text{A} + n_\text{B}}

nA+nB=76nB  ⟹  nA=75nBn_\text{A} + n_\text{B} = 76 n_\text{B} \implies n_\text{A} = 75 n_\text{B}

Substitute nB=115 moln_\text{B} = \frac{1}{15}\text{ mol}: nA=75×115=5 moln_\text{A} = 75 \times \frac{1}{15} = 5\text{ mol}

Hence, the number of moles of the solvent present in the solution is 5.

Calculate Moles of Solvent from Elevation in Boiling Point and Vapour Pressure | Chemistry PYQ Solution - JEE Challenger