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,
[tex]\frac{p^o-p_s}{p^o}=\frac{w_2M_1}{w_1M_2}[/tex]
where,
[tex]p^o[/tex] = vapor pressure of pure solvent (water) = 55.32 mmHg
[tex]p_s[/tex] = vapor pressure of solution = 54.21 mmHg
[tex]w_2[/tex] = mass of solute (urea) = ? g
[tex]w_1[/tex] = mass of solvent (water) = 100 g
[tex]M_1[/tex] = molar mass of solvent (water) = 18 g/mole
[tex]M_2[/tex] = 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.
[tex]\frac{55.32-54.21}{55.32}=\frac{x\times 18}{100\times 60}[/tex]
[tex]x=6.7g[/tex]
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.
