Question

What mass of urea is needed to be dissolved in 200 gm of water to reduce the vapor pressure of water by $50$%?

• A. $333$ g
• B. $111$ g
• C. $50$ g
• D. None of the above

Answer 1

The correct option is A $333$ g

If $p_A^o$ is the vapor pressure of pure water which is taken to be $100$. After the addition of urea the vapor pressure reduced by $50$%. Thus $p_A$ equals to $50$. Hence the relative lowering of vapor pressure is given as $\frac{p_A^o-p_A}{p_A^o}$ that is $\frac{100-50}{100}$$=$$\frac{1}{2}$
Relative lowering of vapor pressure of a solution is equal to the mole fraction of the solute.
Thus $\frac{1}{2}$$=$$X_B$ where $X_B$ is the mole fraction of solute. $X_B$ can also be written as $\frac{W_B}{M_B}\times \frac{M_A}{W_A}$, where $M_A$ and $M_B$ are the molecular mass of solvent (water that is $18$) and solute (urea that is 60) respectively .
$W_B$ and $W_A$ are the mass of solute and solvent present in the solution. $W_B$ is unknown and $W_A$ is $200$.
Putting all the values in the equation $\frac{1}{2}$$=$$\frac{W_B}{M_B}\times \frac{M_A}{W_A}$ we get,
$\frac{1}{2}$$=$$\frac{W_B}{60}\times \frac{18}{200}$
From this the value of $W_B$ is obtained as $333$ g.