Question

The vapour pressure of pure water at ${20}^{o}C$ is $20$ mm $Hg$. A solution of sucrose whose molar mass is $342$ g/mole is prepared by dissolving $75$ g of it in $1000$ g of water. What is the value of relative lowering of vapour pressure of the solution?

• A. $0.0069$
• B. $0.0089$
• C. $0.0039$
• D. $0.0019$

Answer 1

Answer: (C)

$0.0039$

Relative lowering of vapor pressure of the solution is given by the formula $\frac{p_A^o-p_A}{p_A^o}$$=$$X_B$, where $p_A^o$ is the vapor pressure of pure solvent, $p_A$ is the vapor pressure of the solution after adding the solute and $X_B$ is the mole fraction of the solute.
$X_B$ can also be written as $\frac{W_B}{M_B}\times \frac{M_A}{W_A}$
Thus relative lowering of the vapour pressure of the solution is numerically equal to $\frac{W_B}{M_B}\times \frac{M_A}{W_A}$ where $W_B$ and $W_A$ are the amount of solute and solvent present in the solution. $M_B$ and $M_A$ are the molar mass of solute and solvent present in the solution. Putting the values of all the variables in the equation we get the relative lowering of vapor pressure as $\frac{p_A^o-p_A}{p_A^o}$$=$$\frac{75}{342}\times \frac{18}{1000}$

The value obtained is $.0039$ for a relative lowering of vapour pressure.