Respuesta :
Answer : The molecular wight of urea is, 59.9 g/mol
Mole fraction of water and urea is, 0.95 and 0.051 respectively.
Explanation :
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]
