. You prepare a solution by adding 20.0 g Urea [(NH2)2CO] to 125 g water at 25.0oC & vapor pressure 23.76 torr. The solution vapor pressure is 22.67 torr. Calculate the molecular weight of urea. XH2O = moles H2O = Xsolute = moles urea = molar mass = calculations:

According to the relative lowering of vapor pressure, the vapor pressure of a component at a given temperature is equal to the mole fraction of that component of the solution multiplied by the vapor pressure of that component in the pure state.

Formula used :

[tex]\frac{\Delta p}{p^o}=X_b[/tex]

[tex]\frac{p^o-p_s}{p^o}=\frac{n_b}{n_a+n_b}[/tex]

[tex]\frac{p^o-p_s}{p^o}=\frac{\frac{w_b}{M_b}}{\frac{w_a}{M_a}+\frac{w_b}{M_b}}[/tex]

where,

[tex]p^o[/tex] = vapor pressure of the pure solvent water = 23.76 torr

[tex]p_s[/tex] = vapor pressure of the solution = 22.67 torr

[tex]X_b[/tex] = mole fraction of solute (urea)

n is the number of moles

[tex]w_b[/tex] = mass of urea = 20.0 g

[tex]w_a[/tex] = mass of water = 125 g

[tex]M_b[/tex] = molar mass of urea = ?

[tex]M_a[/tex] = molar mass of water = 18 g/mol

Now put all the given values in the above expression, we get:

[tex]\frac{23.76-22.67}{23.76}=\frac{\frac{20.0}{M_b}}{\frac{125}{18}+\frac{20.0}{M_b}}[/tex]

[tex]M_b=59.9g/mol[/tex]

Now we have to calculate the moles of [tex]H_2O[/tex] and urea.

[tex]\text{Moles of }H_2O=\frac{\text{Given mass }H_2O}{\text{Molar mass }H_2O}=\frac{125g}{18g/mol}=6.94mol[/tex]

and,

[tex]\text{Moles of urea}=\frac{\text{Given mass urea}}{\text{Molar mass urea}}=\frac{20.0g}{59.9g/mol}=0.334mol[/tex]

Now we have to calculate the mole fraction of [tex]H_2O[/tex] and urea.

[tex]X_{H_2O}=\frac{n_{H_2O}}{n_{H_2O}+n_{urea}}[/tex]

[tex]X_{H_2O}=\frac{6.94}{6.94+0.334}[/tex]

[tex]X_{H_2O}=0.95[/tex]

and,

[tex]X_{urea}=\frac{n_{urea}}{n_{H_2O}+n_{urea}}[/tex]

[tex]X_{urea}=\frac{0.334}{6.94+0.334}[/tex]

[tex]X_{urea}=0.051[/tex]