Question

The Vapour pressure of a solution of 5g of non-electrolyte in 100g of water at a particular temperature is 2985 $N{{m}^{-2}}$. The vapour pressure of pure water at that temperature is 3000 $N{{m}^{-2}}$. The molecular weight of solute is

Answer 1

Vapour pressure of pure water will be lowered when a solute is added. As more solute is added, vapour pressure goes on lowering. Relative lowering of vapour pressure is the ratio of lowering of vapour pressure to vapour pressure of pure component.

Complete step by step solution:
-Raoult’s Law states that partial vapour pressure of each component is directly proportional to its mole fraction present in the component.
${{P}_{1}}^{0}$is vapour pressure of pure component 1

& {{P}_{2}}\propto {{\text{x}}_{2}} \\\ & ={{P}_{2}}={{P}_{2}}^{0}{{x}_{2}} \\\ \end{aligned}$$ ${{P}_{2}}^{0}$is vapour pressure of pure component 2 $$\begin{aligned} & {{P}_{T}}={{P}_{1}}+{{P}_{2}} \\\ & ={{P}_{1}}^{0}.{{x}_{1}}+{{P}_{2}}^{0}.{{x}_{2}} \\\ & ={{P}_{1}}^{0}.{{x}_{1}}+{{P}_{2}}^{0}(1-{{x}_{1}}) \\\ & ={{P}_{1}}^{0}.{{x}_{1}}+{{P}_{2}}^{0}-{{P}_{2}}^{0}.{{x}_{1}} \\\ & \\\ \end{aligned}$$ As Solute is nonvolatile, it will not exert any vapour pressure, so vapour pressure of component 2 is zero, hence ${{P}_{1}}={{P}_{1}}^{0}.{{x}_{1}}$ Lowering of vapour pressure is the difference between vapour pressure of pure components and vapour pressure of solution. $$\Delta \text{P= }{{\text{P}}_{1}}^{0}-P$$ $={{P}_{1}}^{0}-{{P}_{1}}^{0}.{{X}_{1}}={{P}_{1}}^{0}(1-{{x}_{2}})={{P}_{1}}^{0}.{{x}_{2}}$ Relative lowering of vapour pressure is ratio of lowering of vapour pressure to vapour pressure of pure component. Relative lowering of vapour pressure=$\dfrac{{{P}_{1}}^{0}.{{x}_{2}}}{{{P}_{1}}^{0}}={{x}_{2}}$ $${{x}_{2}}$$is mole fraction of solution which is ratio of number of moles of solute to sum of moles of all components present in solution. $${{x}_{2}}=\dfrac{{{n}_{2}}}{{{n}_{1}}+{{n}_{2}}}$$ Number of moles of solute can be neglected compared to the number of moles of solvent. $$\dfrac{\Delta P}{{{P}^{0}}_{1}}=\dfrac{{{P}^{0}}_{1}-P}{{{P}^{0}}_{1}}=\dfrac{3000-2985}{3000}$$=0.005 $\begin{aligned} & {{x}_{2}}=\dfrac{{{w}_{2}}{{M}_{1}}}{{{M}_{2}}{{w}_{1}}} \\\ & 0.005=\dfrac{5\times 18}{{{M}_{2}}\times 100} \\\ & {{M}_{2}}=\dfrac{90}{0.5}=180 \\\ \end{aligned}$ **Molecular weight of solute is 180g/mol.** **Note:** Relative lowering of vapour pressure can be determined by calculating mole fraction of solute. When mole of solute is very small compared to moles of the solvent, the mole of solute can be neglected. Lowering of vapour pressure can be calculated by taking the difference between the vapour pressure of pure components and solution.