(B) According to Raoult's Law for relative lowering of vapour pressure:$\frac{P^0 - P_s}{P_s} = \frac{n_2}{n_1} = \frac{w_2 \times M_1}{M_2 \times w_1

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(B) According to Raoult's Law for relative lowering of vapour pressure:$\frac{P^0 - P_s}{P_s} = \frac{n_2}{n_1} = \frac{w_2 \times M_1}{M_2 \times w_1}$Given:$P^0 = 640 \ mm \ Hg$,$P_s = 600 \ mm \ Hg$,$w_2 = 2.175 \ g$,$w_1 = 39.0 \ g$,$M_1 (\text{benzene, } C_6H_6) = 78 \ g/mol$Substituting the values:$\frac{640 - 600}{600} = \frac{2.175 \times 78}{M_2 \times 39.0}$$\frac{40}{600} = \frac{2.175 \times 2}{M_2}$$\frac{1}{15} = \frac{4.35}{M_2}$$M_2 = 4.35 \times 15 = 65.25 \ 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 $640 \ mm \ Hg$. $A$ non-volatile non-electrolyte solid weighing $2.175 \ g$ is added to $39.0 \ g$ of benzene. The vapour pressure of the solution is $600 \ mm \ Hg$. What is the molar mass of the solid substance?

A
$6.96$
B
$65.25$
C
$63.8$
D
None of these

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Lowering of vapour pressure

$1 \ mol$ of heptane $(V.P = 92 \ mm \ Hg)$ is mixed with $4 \ mol$ of octane $(V.P = 31 \ mm \ Hg)$. The vapor pressure of the resulting ideal solution is .......... $mm \ Hg$.

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The vapour pressure of a solution of methyl alcohol and ethyl alcohol is represented as $P = 115 X_A + 140$,where $X_A$ is the mole fraction of methyl alcohol. The value of $\lim_{X_A \to 0} \frac{P_B^o}{X_B}$ is

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At a given temperature,the vapour pressure of a solution of two volatile liquids $A$ and $B$ is given by the equation $P_S = 120 - 80 \, X_B$ (where $X_B$ is the mole fraction of $B$). The vapour pressures of pure $A$ and $B$ at the same temperature are respectively:

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The vapour pressure of water at room temperature is $23.8 \ mm \ Hg$. The vapour pressure of an aqueous solution of sucrose with mole fraction of $0.1$ is equal to .......... $mm \ Hg$.

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The relative lowering in vapour pressure of an aqueous solution of urea which is $6 \%$ by weight,is

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The vapour pressure of pure benzene at a certain temperature is $640 \ mm \ Hg$. $A$ non-volatile non-electrolyte solid weighing $2.175 \ g$ is added to $39.0 \ g$ of benzene. The vapour pressure of the solution is $600 \ mm \ Hg$. What is the molar mass of the solid substance?

Similar Questions

$1 \ mol$ of heptane $(V.P = 92 \ mm \ Hg)$ is mixed with $4 \ mol$ of octane $(V.P = 31 \ mm \ Hg)$. The vapor pressure of the resulting ideal solution is .......... $mm \ Hg$.

The vapour pressure of a solution of methyl alcohol and ethyl alcohol is represented as $P = 115 X_A + 140$,where $X_A$ is the mole fraction of methyl alcohol. The value of $\lim_{X_A \to 0} \frac{P_B^o}{X_B}$ is

At a given temperature,the vapour pressure of a solution of two volatile liquids $A$ and $B$ is given by the equation $P_S = 120 - 80 \, X_B$ (where $X_B$ is the mole fraction of $B$). The vapour pressures of pure $A$ and $B$ at the same temperature are respectively:

The vapour pressure of water at room temperature is $23.8 \ mm \ Hg$. The vapour pressure of an aqueous solution of sucrose with mole fraction of $0.1$ is equal to .......... $mm \ Hg$.

The relative lowering in vapour pressure of an aqueous solution of urea which is $6 \%$ by weight,is

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