Question

3g of urea is added to 36 g of boiling water. How much lowering its vapour pressure is noticed

Answer 1

The vapour pressure is found out to be 0.975 atm.

Answer 2

Answer:  18.24 mm Hg .

Explanation:

As the relative lowering of vapor pressure is directly proportional to the amount of dissolved solute.

The formula for relative lowering of vapor pressure will be,

$\frac{p^o-p_s}{p^o}=x_B$

where,

$p^o$ = vapor pressure of pure solvent  (water) = 760 mm Hg

$p_s$ = vapor pressure of solution  = ?

$x_B$ = mole fraction of solute (urea) = ?

$\frac{760-p_s}{760}=x_B$

We have to calculate the mole fraction of solute (urea)

$x_{B}=(\frac{n_{B}}{n_{A}+n_{B}})=\frac{(\frac{w_B}{M_B})}{(\frac{w_A}{M_A})+(\frac{w_B}{M_B})}$

$x_{B}=\frac{(\frac{3}{60})}{(\frac{3}{60})+(\frac{36}{18})}$

$x_B=0.024$

Putting in the values, we get:

$\frac{760-p_s}{760}=0.024$

$p_s=741.76mm Hg$

Lowering in vapour pressure is = 760 - 741.76 = 18.24 mm Hg

Thus lowering in vapour pressure is 18.24 mm Hg .