(A) Step $1$: Identify the formula for vapour pressure of a solution using Raoult's Law: $P_s = P^0 \times x_1$, where $P_s$ is the vapour pressure of

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(A) Step $1$: Identify the formula for vapour pressure of a solution using Raoult's Law: $P_s = P^0 \times x_1$, where $P_s$ is the vapour pressure of the solution, $P^0$ is the vapour pressure of pure solvent, and $x_1$ is the mole fraction of the solvent.Step $2$: Substitute the given values into the formula: $P^0 = 18 \text{ mm Hg}$ and $x_1 = 0.9964$.Step $3$: Calculate the result: $P_s = 18 \text{ mm Hg} \times 0.9964 = 17.9352 \text{ mm Hg}$.Step $4$: Rounding to two decimal places, we get $P_s \approx 17.94 \text{ mm Hg}$.

Class 12 Chemistry Solutions Lowering of vapour pressure

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$A$ solution is prepared by dissolving $68 \text{ g}$ of sucrose $(C_{12}H_{22}O_{11})$ in $1 \text{ kg}$ of water. Calculate the vapour pressure of the solution at $298 \text{ K}$. Given: Vapour pressure of pure water at $298 \text{ K} = 18 \text{ mm Hg}$ and mole fraction of solvent $(x_1)$ $= 0.9964$. (in $\text{ mm Hg}$)

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A
$17.94$
B
$18.06$
C
$17.50$
D
$18.50$

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Lowering of vapour pressure

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If $8 \; g$ of a non-electrolyte solute is dissolved in $114 \; g$ of $n$-octane to reduce its vapour pressure to $80 \%$,the molar mass (in $g \; mol^{-1}$) of the solute is.
Given that the molar mass of $n$-octane is $114 \; g \; mol^{-1}$.

MediumNEET 2020

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Calculate the vapour pressure of pure volatile liquid $A$ at a given temperature if the mole fraction and vapour pressure of pure volatile liquid $B$ are $0.4$ and $900 \ mm \ Hg$ respectively,given that the total vapour pressure of the solution is $600 \ mm \ Hg$. (in $mm \ Hg$)

MediumMHT CET 2025

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$A$ solution containing $30 \, g$ of non-volatile solute in exactly $90 \, g$ water has a vapour pressure of $21.85 \, mm \, Hg$ at $25 \, ^oC$. Further $18 \, g$ of water is then added to the solution. The resulting solution has a vapour pressure of $22.15 \, mm \, Hg$ at $25 \, ^oC$. Calculate the molecular weight of the solute.

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"The relative lowering of the vapour pressure is equal to the mole fraction of the solute." This law is called

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At $50^{\circ} C$, the vapour pressure of pure benzene is $268 \ torr$. The number of moles of non-volatile solute per mole of benzene required to prepare a solution having a vapour pressure of $167 \ torr$ at the same temperature is (molar mass of benzene $= 78 \ g \ mol^{-1}$)

MediumTS EAMCET 2023

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If $8 \; g$ of a non-electrolyte solute is dissolved in $114 \; g$ of $n$-octane to reduce its vapour pressure to $80 \%$,the molar mass (in $g \; mol^{-1}$) of the solute is.Given that the molar mass of $n$-octane is $114 \; g \; mol^{-1}$.

Calculate the vapour pressure of pure volatile liquid $A$ at a given temperature if the mole fraction and vapour pressure of pure volatile liquid $B$ are $0.4$ and $900 \ mm \ Hg$ respectively,given that the total vapour pressure of the solution is $600 \ mm \ Hg$. (in $mm \ Hg$)

"The relative lowering of the vapour pressure is equal to the mole fraction of the solute." This law is called

At $50^{\circ} C$, the vapour pressure of pure benzene is $268 \ torr$. The number of moles of non-volatile solute per mole of benzene required to prepare a solution having a vapour pressure of $167 \ torr$ at the same temperature is (molar mass of benzene $= 78 \ g \ mol^{-1}$)

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If $8 \; g$ of a non-electrolyte solute is dissolved in $114 \; g$ of $n$-octane to reduce its vapour pressure to $80 \%$,the molar mass (in $g \; mol^{-1}$) of the solute is.
Given that the molar mass of $n$-octane is $114 \; g \; mol^{-1}$.