(D) Given mole ratio of pentane $(n_p)$ to hexane $(n_h)$ is $1:4$. Mole fraction of pentane $(x_p)$ = $\frac{1}{1+4} = 0.2$. Mole fraction of hexane

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(D) Given mole ratio of pentane $(n_p)$ to hexane $(n_h)$ is $1:4$. Mole fraction of pentane $(x_p)$ = $\frac{1}{1+4} = 0.2$. Mole fraction of hexane $(x_h)$ = $\frac{4}{1+4} = 0.8$. Total vapour pressure $(P_T)$ = $P_p^0 x_p + P_h^0 x_h$. $P_T = (440 \times 0.2) + (120 \times 0.8) = 88 + 96 = 184 \ mm \ Hg$. According to Dalton's law,the mole fraction of pentane in the vapour phase $(y_p)$ is given by $y_p = \frac{P_p^0 x_p}{P_T}$. $y_p = \frac{88}{184} \approx 0.478$.

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$A$ solution has a $1:4$ mole ratio of pentane to hexane. The vapour pressure of the pure hydrocarbons at $20 \ ^oC$ are $440 \ mm \ Hg$ for pentane and $120 \ mm \ Hg$ for hexane. The mole fraction of pentane in the vapour phase would be

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A
$0.549$
B
$0.2$
C
$0.786$
D
$0.478$

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At $80\,^oC$,the vapour pressure of pure liquid $A$ is $520\, mm\,Hg$ and that of pure liquid $B$ is $1000\, mm\,Hg$. If a mixture of $A$ and $B$ boils at $80\,^oC$ and $1\, atm$ pressure,the amount of $A$ in the mixture is ........... $mol\,\%$.

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Statement $1$: The cold pack used by athletes to treat injuries on the field works by dissolving a solute in water,which causes the temperature to decrease. Statement $2$: Adding a non-volatile solute to a solvent decreases its freezing point.

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An aqueous solution of a non-volatile solute boils at $100.17^{\circ} C$. The temperature at which this solution will freeze (in $^{\circ} C$) is
$K_{b}(H_2 O) = 0.512^{\circ} C \ kg \ mol^{-1}$,
$K_{f}(H_2 O) = 1.86^{\circ} C \ kg \ mol^{-1}$

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The only incorrect statement regarding Henry's law for a gas dissolving in water is

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Assertion : If one component of a solution obeys Raoult's law over a certain range of composition,the other component will not obey Henry's law in that range.
Reason : Raoult's law is a special case of Henry's law.

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$A$ solution has a $1:4$ mole ratio of pentane to hexane. The vapour pressure of the pure hydrocarbons at $20 \ ^oC$ are $440 \ mm \ Hg$ for pentane and $120 \ mm \ Hg$ for hexane. The mole fraction of pentane in the vapour phase would be

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At $80\,^oC$,the vapour pressure of pure liquid $A$ is $520\, mm\,Hg$ and that of pure liquid $B$ is $1000\, mm\,Hg$. If a mixture of $A$ and $B$ boils at $80\,^oC$ and $1\, atm$ pressure,the amount of $A$ in the mixture is ........... $mol\,\%$.

Statement $1$: The cold pack used by athletes to treat injuries on the field works by dissolving a solute in water,which causes the temperature to decrease. Statement $2$: Adding a non-volatile solute to a solvent decreases its freezing point.

An aqueous solution of a non-volatile solute boils at $100.17^{\circ} C$. The temperature at which this solution will freeze (in $^{\circ} C$) is$K_{b}(H_2 O) = 0.512^{\circ} C \ kg \ mol^{-1}$,$K_{f}(H_2 O) = 1.86^{\circ} C \ kg \ mol^{-1}$

The only incorrect statement regarding Henry's law for a gas dissolving in water is

Assertion : If one component of a solution obeys Raoult's law over a certain range of composition,the other component will not obey Henry's law in that range.Reason : Raoult's law is a special case of Henry's law.

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An aqueous solution of a non-volatile solute boils at $100.17^{\circ} C$. The temperature at which this solution will freeze (in $^{\circ} C$) is
$K_{b}(H_2 O) = 0.512^{\circ} C \ kg \ mol^{-1}$,
$K_{f}(H_2 O) = 1.86^{\circ} C \ kg \ mol^{-1}$

Assertion : If one component of a solution obeys Raoult's law over a certain range of composition,the other component will not obey Henry's law in that range.
Reason : Raoult's law is a special case of Henry's law.