Q.75. A non-volatile, non-electrolyte solid solute when dissolved in 40 g of a solvent, the vapour p

Q: What is the correct answer for this exam Chemistry Q75?

The correct answer is 5. Sol. $\frac{P^0 - P_s}{P_s}$ = i. molality $\times \frac{\text{(M.solvent)}}{1000}$ $\Delta T_b$ = i. K$_b$. molality $\Rightarrow$ molality = $\frac{0.5}{0.3}$ (Molecular Mass) = $\frac{600}{75}$ g Moles = $\frac{40}{600/75} = 5$

Q: Q.75. A non-volatile, non-electrolyte solid solute when dissolved in 40 g of a solvent, the vapour pressure of the solvent decreased from 760 mm Hg to 750 mm Hg. If the same solution boils at 320 K,…

The correct answer is 5. Sol. $\frac{P^0 - P_s}{P_s}$ = i. molality $\times \frac{\text{(M.solvent)}}{1000}$ $\Delta T_b$ = i. K$_b$. molality $\Rightarrow$ molality = $\frac{0.5}{0.3}$ (Molecular Mass) = $\frac{600}{75}$ g Moles = $\frac{40}{600/75} = 5$

Question

Q.75. A non-volatile, non-electrolyte solid solute when dissolved in 40 g of a solvent, the vapour pressure of the solvent decreased from 760 mm Hg to 750 mm Hg. If the same solution boils at 320 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, K of the solvent = 0.3 K kg mol⁻¹]

FirstInTest · Accepted Answer

Sol. $\frac{P^0 - P_s}{P_s}$ = i. molality $	imes \frac{	ext{(M.solvent)}}{1000}$ $\Delta T_b$ = i. K$_b$. molality $\Rightarrow$ molality = $\frac{0.5}{0.3}$ (Molecular Mass) = $\frac{600}{75}$ g Moles = $\frac{40}{600/75} = 5$

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