Question

What mass (in g) of urea (CO(NH2)2) in 100.0 g of water is needed to decrease the vapor pressure of water from 55.32 mmHg of pure water to 54.21 mmHg for the solution? The temperature is held constant at 40oC.

Answer 1

Answer: 6.7 g

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,

$\frac{p^o-p_s}{p^o}=\frac{w_2M_1}{w_1M_2}$

where,

$p^o$ = vapor pressure of pure solvent  (water) = 55.32 mmHg

$p_s$ = vapor pressure of solution = 54.21 mmHg

$w_2$ = mass of solute  (urea) = ? g

$w_1$ = mass of solvent  (water) = 100 g

$M_1$ = molar mass of solvent (water) = 18 g/mole

$M_2$ = molar mass of solute (urea) = 60 g/mole

Now put all the given values in this formula ,we get the vapor pressure of the solution.

$\frac{55.32-54.21}{55.32}=\frac{x\times 18}{100\times 60}$

$x=6.7g$

Therefore, 6.7 g of urea is needed in 100.0 g of water is needed to decrease the vapor pressure of water from 55.32 mmHg of pure water to 54.21 mmHg for the solution.