277301 The vapour pressure of a solvent decreased by $10 \mathrm{~mm}$ of mercury when a non-volatile solute was added to the solvent. The mol fraction of the solute in the solution is 0.2 . What should be the mole fraction of the solvent if the decrease in the vapour pressure is to be $20 \mathrm{~mm}$ of mercury ?

277305 The freezing point of an aqueous solution containing $25 \mathrm{~g}$ of ethanol in $1000 \mathrm{~g}$ of $\mathrm{H}_{2} \mathrm{O}$ is $\left[K_{\mathrm{f}}=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right]$

277306 The vapour pressure of pure liquids $A$ and $B$ are 450 and $700 \mathrm{~mm}$ at $\mathrm{Hg}$ of $350 \mathrm{~K}$ respectively. If the total vapour pressure of the mixture is $600 \mathrm{~mm}$ of $\mathrm{Hg}$, the composition of the mixture in the solution is

277307 Henry's law constant for the solubility of $\mathrm{N}_{2}$ gas in water at $298 \mathrm{~K}$ is $1.0 \times 10^{5} \mathrm{~atm}$. The mole fraction of $\mathrm{N}_{2}$ in air is 0.8 . The number of moles of $\mathrm{N}_{2}$ from air dissolved in 10 moles of water at $298 \mathrm{~K}$ and $5 \mathbf{a t m}$ pressure is

277308 The correct option for the value of vapour pressure of a solution at $45^{\circ} \mathrm{C}$ with benzene to octane in molar ratio $3: 2$ is
[At $45^{\circ} \mathrm{C}$ pressure of benzene is $280 \mathrm{~mm} \mathrm{Hg}$ and that of octance is $420 \mathrm{~mm} \mathrm{Hg}$. Assume ideal gas]

277301 The vapour pressure of a solvent decreased by $10 \mathrm{~mm}$ of mercury when a non-volatile solute was added to the solvent. The mol fraction of the solute in the solution is 0.2 . What should be the mole fraction of the solvent if the decrease in the vapour pressure is to be $20 \mathrm{~mm}$ of mercury ?

277305 The freezing point of an aqueous solution containing $25 \mathrm{~g}$ of ethanol in $1000 \mathrm{~g}$ of $\mathrm{H}_{2} \mathrm{O}$ is $\left[K_{\mathrm{f}}=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right]$

277306 The vapour pressure of pure liquids $A$ and $B$ are 450 and $700 \mathrm{~mm}$ at $\mathrm{Hg}$ of $350 \mathrm{~K}$ respectively. If the total vapour pressure of the mixture is $600 \mathrm{~mm}$ of $\mathrm{Hg}$, the composition of the mixture in the solution is

277307 Henry's law constant for the solubility of $\mathrm{N}_{2}$ gas in water at $298 \mathrm{~K}$ is $1.0 \times 10^{5} \mathrm{~atm}$. The mole fraction of $\mathrm{N}_{2}$ in air is 0.8 . The number of moles of $\mathrm{N}_{2}$ from air dissolved in 10 moles of water at $298 \mathrm{~K}$ and $5 \mathbf{a t m}$ pressure is

277308 The correct option for the value of vapour pressure of a solution at $45^{\circ} \mathrm{C}$ with benzene to octane in molar ratio $3: 2$ is
[At $45^{\circ} \mathrm{C}$ pressure of benzene is $280 \mathrm{~mm} \mathrm{Hg}$ and that of octance is $420 \mathrm{~mm} \mathrm{Hg}$. Assume ideal gas]

277301 The vapour pressure of a solvent decreased by $10 \mathrm{~mm}$ of mercury when a non-volatile solute was added to the solvent. The mol fraction of the solute in the solution is 0.2 . What should be the mole fraction of the solvent if the decrease in the vapour pressure is to be $20 \mathrm{~mm}$ of mercury ?

277305 The freezing point of an aqueous solution containing $25 \mathrm{~g}$ of ethanol in $1000 \mathrm{~g}$ of $\mathrm{H}_{2} \mathrm{O}$ is $\left[K_{\mathrm{f}}=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right]$

277306 The vapour pressure of pure liquids $A$ and $B$ are 450 and $700 \mathrm{~mm}$ at $\mathrm{Hg}$ of $350 \mathrm{~K}$ respectively. If the total vapour pressure of the mixture is $600 \mathrm{~mm}$ of $\mathrm{Hg}$, the composition of the mixture in the solution is

277307 Henry's law constant for the solubility of $\mathrm{N}_{2}$ gas in water at $298 \mathrm{~K}$ is $1.0 \times 10^{5} \mathrm{~atm}$. The mole fraction of $\mathrm{N}_{2}$ in air is 0.8 . The number of moles of $\mathrm{N}_{2}$ from air dissolved in 10 moles of water at $298 \mathrm{~K}$ and $5 \mathbf{a t m}$ pressure is

277308 The correct option for the value of vapour pressure of a solution at $45^{\circ} \mathrm{C}$ with benzene to octane in molar ratio $3: 2$ is
[At $45^{\circ} \mathrm{C}$ pressure of benzene is $280 \mathrm{~mm} \mathrm{Hg}$ and that of octance is $420 \mathrm{~mm} \mathrm{Hg}$. Assume ideal gas]

277301 The vapour pressure of a solvent decreased by $10 \mathrm{~mm}$ of mercury when a non-volatile solute was added to the solvent. The mol fraction of the solute in the solution is 0.2 . What should be the mole fraction of the solvent if the decrease in the vapour pressure is to be $20 \mathrm{~mm}$ of mercury ?

277305 The freezing point of an aqueous solution containing $25 \mathrm{~g}$ of ethanol in $1000 \mathrm{~g}$ of $\mathrm{H}_{2} \mathrm{O}$ is $\left[K_{\mathrm{f}}=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right]$

277306 The vapour pressure of pure liquids $A$ and $B$ are 450 and $700 \mathrm{~mm}$ at $\mathrm{Hg}$ of $350 \mathrm{~K}$ respectively. If the total vapour pressure of the mixture is $600 \mathrm{~mm}$ of $\mathrm{Hg}$, the composition of the mixture in the solution is

277307 Henry's law constant for the solubility of $\mathrm{N}_{2}$ gas in water at $298 \mathrm{~K}$ is $1.0 \times 10^{5} \mathrm{~atm}$. The mole fraction of $\mathrm{N}_{2}$ in air is 0.8 . The number of moles of $\mathrm{N}_{2}$ from air dissolved in 10 moles of water at $298 \mathrm{~K}$ and $5 \mathbf{a t m}$ pressure is

277308 The correct option for the value of vapour pressure of a solution at $45^{\circ} \mathrm{C}$ with benzene to octane in molar ratio $3: 2$ is
[At $45^{\circ} \mathrm{C}$ pressure of benzene is $280 \mathrm{~mm} \mathrm{Hg}$ and that of octance is $420 \mathrm{~mm} \mathrm{Hg}$. Assume ideal gas]

277301 The vapour pressure of a solvent decreased by $10 \mathrm{~mm}$ of mercury when a non-volatile solute was added to the solvent. The mol fraction of the solute in the solution is 0.2 . What should be the mole fraction of the solvent if the decrease in the vapour pressure is to be $20 \mathrm{~mm}$ of mercury ?

277305 The freezing point of an aqueous solution containing $25 \mathrm{~g}$ of ethanol in $1000 \mathrm{~g}$ of $\mathrm{H}_{2} \mathrm{O}$ is $\left[K_{\mathrm{f}}=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right]$

277306 The vapour pressure of pure liquids $A$ and $B$ are 450 and $700 \mathrm{~mm}$ at $\mathrm{Hg}$ of $350 \mathrm{~K}$ respectively. If the total vapour pressure of the mixture is $600 \mathrm{~mm}$ of $\mathrm{Hg}$, the composition of the mixture in the solution is

277307 Henry's law constant for the solubility of $\mathrm{N}_{2}$ gas in water at $298 \mathrm{~K}$ is $1.0 \times 10^{5} \mathrm{~atm}$. The mole fraction of $\mathrm{N}_{2}$ in air is 0.8 . The number of moles of $\mathrm{N}_{2}$ from air dissolved in 10 moles of water at $298 \mathrm{~K}$ and $5 \mathbf{a t m}$ pressure is

277308 The correct option for the value of vapour pressure of a solution at $45^{\circ} \mathrm{C}$ with benzene to octane in molar ratio $3: 2$ is
[At $45^{\circ} \mathrm{C}$ pressure of benzene is $280 \mathrm{~mm} \mathrm{Hg}$ and that of octance is $420 \mathrm{~mm} \mathrm{Hg}$. Assume ideal gas]