At 300 K,the vapour pressure of a solution c | vedclass

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(A) Let $P_{A}^{\circ}$ be the vapour pressure of pure $n-hexane$ and $P_{B}^{\circ}$ be the vapour pressure of pure $n-heptane$.According to Raoult's

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$600$

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(A) Let $P_{A}^{\circ}$ be the vapour pressure of pure $n-hexane$ and $P_{B}^{\circ}$ be the vapour pressure of pure $n-heptane$.According to Raoult's Law,$P_{total} = X_{A}P_{A}^{\circ} + X_{B}P_{B}^{\circ}$.For the first case: $X_{A} = \frac{1}{1+3} = \frac{1}{4}$ and $X_{B} = \frac{3}{4}$.$550 = \frac{1}{4}P_{A}^{\circ} + \frac{3}{4}P_{B}^{\circ} \implies P_{A}^{\circ} + 3P_{B}^{\circ} = 2200 \dots (i)$For the second case: $1 \ mole$ of $n-heptane$ is added,so total moles = $1 + 4 = 5$. $X_{A} = \frac{1}{5}$ and $X_{B} = \frac{4}{5}$.The new vapour pressure is $550 + 10 = 560 \ mm \ Hg$.$560 = \frac{1}{5}P_{A}^{\circ} + \frac{4}{5}P_{B}^{\circ} \implies P_{A}^{\circ} + 4P_{B}^{\circ} = 2800 \dots (ii)$Subtracting equation $(i)$ from $(ii)$:$(P_{A}^{\circ} + 4P_{B}^{\circ}) - (P_{A}^{\circ} + 3P_{B}^{\circ}) = 2800 - 2200$$P_{B}^{\circ} = 600 \ mm \ Hg$.

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When $25 \ g$ of a non-volatile solute is dissolved in $100 \ g$ of water, the vapour pressure is lowered by $2.25 \times 10^{-1} \ mm$. If the vapour pressure of water at $20^{\circ}C$ is $17.5 \ mm$, what is the molecular weight of the solute?

Which one of the following is the expression of Raoult's law?
$p$ = vapour pressure of pure solvent
$p_s$ = vapour pressure of the solution
$n$ = number of moles of the solute
$N$ = number of moles of the solvent

Which solution will show the maximum vapour pressure at $300 \ K$?

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