The vapor pressure of pure A is 10 torr. Whe | vedclass

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(A) According to Raoult's Law for a non-volatile solute,the relative lowering of vapor pressure is given by: $\frac{P^o_A - P_s}{P^o_A} = \frac{n_B}{n

Vedclass · Accepted Answer

$100$

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(A) According to Raoult's Law for a non-volatile solute,the relative lowering of vapor pressure is given by: $\frac{P^o_A - P_s}{P^o_A} = \frac{n_B}{n_A + n_B} \approx \frac{n_B}{n_A}$.Given: $P^o_A = 10 \ torr$,$P_s = 9 \ torr$,$w_B = 1 \ g$,$w_A = 20 \ g$,$M_A = 200 \ g/mol$.Substituting the values: $\frac{10 - 9}{10} = \frac{w_B / M_B}{w_A / M_A}$.$0.1 = \frac{1 / M_B}{20 / 200}$.$0.1 = \frac{1 / M_B}{0.1}$.$0.1 	imes 0.1 = \frac{1}{M_B}$.$0.01 = \frac{1}{M_B}$.$M_B = \frac{1}{0.01} = 100 \ g/mol$.

The vapor pressure of pure $A$ is $10 \ torr$. When $1 \ g$ of $B$ is dissolved in $20 \ g$ of $A$ at the same temperature,the vapor pressure of the solution becomes $9 \ torr$. If the molar mass of $A$ is $200 \ g/mol$,the molar mass of $B$ is ............ $amu$.

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