(D) Given: $P^o = 200 \ mm \ Hg$ $P_s = 195 \ mm \ Hg$ $w = 2 \ g$ (mass of solute) $W = 78 \ g$ (mass of solvent,benzene) $M = 78 \ g/mol$ (molar mas

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(D) Given: $P^o = 200 \ mm \ Hg$ $P_s = 195 \ mm \ Hg$ $w = 2 \ g$ (mass of solute) $W = 78 \ g$ (mass of solvent,benzene) $M = 78 \ g/mol$ (molar mass of benzene,$C_6H_6$) Using Raoult's law for non-volatile solute: $\frac{P^o - P_s}{P^o} = \frac{w \times M}{m \times W}$ $\frac{200 - 195}{200} = \frac{2 \times 78}{m \times 78}$ $\frac{5}{200} = \frac{2}{m}$ $m = \frac{2 \times 200}{5} = 80 \ g/mol$ Therefore,the molecular weight of the solid is $80 \ g/mol$.

Class 12 Chemistry Solutions Lowering of vapour pressure

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The vapour pressure of pure benzene at a certain temperature is $200 \ mm \ Hg$. At the same temperature,the vapour pressure of a solution containing $2 \ g$ of a non-volatile,non-electrolyte solid in $78 \ g$ of benzene is $195 \ mm \ Hg$. What is the molecular weight of the solid?

A
$50$
B
$70$
C
$85$
D
$80$

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Lowering of vapour pressure

When $5 \ g$ of a non-electrolyte solute is dissolved in $100 \ g$ of water,the vapor pressure decreases from $3000 \ N \ m^{-2}$ to $2985 \ N \ m^{-2}$. The molar mass of the solute is ............... $g/mol$.

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Calculate the weight of non-volatile solute having molecular weight $40$,which is dissolved in $57 \ g$ octane to reduce its vapour pressure by $20\%$.

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Vapour pressure of a solution containing $18 \ g$ of glucose and $178.2 \ g$ of water at $100^{\circ} C$ is (Vapour pressure of pure water at $100^{\circ} C = 760 \ torr$) (in $torr$)

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State Raoult's law for a solution containing a non-volatile solute and a volatile solvent and explain it.

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At a certain temperature,$5 \ g$ of a non-electrolyte solute is dissolved in $100 \ g$ of water. The vapour pressure of the resulting solution is $2985 \ N/m^2$,and the vapour pressure of pure water is $3000 \ N/m^2$. Calculate the molar mass of the solute.

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The vapour pressure of pure benzene at a certain temperature is $200 \ mm \ Hg$. At the same temperature,the vapour pressure of a solution containing $2 \ g$ of a non-volatile,non-electrolyte solid in $78 \ g$ of benzene is $195 \ mm \ Hg$. What is the molecular weight of the solid?

Similar Questions

When $5 \ g$ of a non-electrolyte solute is dissolved in $100 \ g$ of water,the vapor pressure decreases from $3000 \ N \ m^{-2}$ to $2985 \ N \ m^{-2}$. The molar mass of the solute is ............... $g/mol$.

Calculate the weight of non-volatile solute having molecular weight $40$,which is dissolved in $57 \ g$ octane to reduce its vapour pressure by $20\%$.

Vapour pressure of a solution containing $18 \ g$ of glucose and $178.2 \ g$ of water at $100^{\circ} C$ is (Vapour pressure of pure water at $100^{\circ} C = 760 \ torr$) (in $torr$)

State Raoult's law for a solution containing a non-volatile solute and a volatile solvent and explain it.

At a certain temperature,$5 \ g$ of a non-electrolyte solute is dissolved in $100 \ g$ of water. The vapour pressure of the resulting solution is $2985 \ N/m^2$,and the vapour pressure of pure water is $3000 \ N/m^2$. Calculate the molar mass of the solute.

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