Respuesta :
First, we have to remember that the vapor pressure is the pressure that a gas phase exerts on the liquid phase. It changes with the temperature and the change in the concentration of a solute.
The equation to calculate the change in the vapor pressure of a solution is:
[tex]\begin{gathered} \Delta P=\text{ x}_2*P_1^o \\ \\ \Delta P\text{ is the change in the vapor pressure} \\ x_2\text{ is the molar fraction os the solute} \\ P_1^o\text{ is the initial vapor pressure of the solvent. } \end{gathered}[/tex]We have the initial vapor pressure and we have the molality of the urea solution. Then, we have to convert the molality into a molar fraction, with the help of the molecular weight of the water.
[tex]\begin{gathered} m=\text{ 1.00}\frac{moles\text{ urea}}{kg\text{ water}} \\ \\ 1.00\text{ }\frac{moles\text{ urea}}{kg\text{ water}}*\frac{18\text{ kg water}}{1\text{ kmol water}}*\frac{1\text{ kmol water}}{1000\text{ mol water}}=0.018\text{ }\frac{mol\text{ urea}}{mol\text{ water}} \end{gathered}[/tex][tex]0.018\text{ }\frac{mol\text{ urea}}{mol\text{ water}}\rightarrow\text{ x}_2=\text{ }\frac{0.018}{1+0.018}=\text{ 0.01768}[/tex]Then, we have all the necessary data to calculate the vapor pressure change of the solution:
[tex]\begin{gathered} \Delta P=0.01768\text{ * 17.5 mmHg = 0.3094 mmHg} \\ \end{gathered}[/tex]Finally, we can calculate the final vapor pressure of the solution:
[tex]\begin{gathered} P^o=P_1^o\text{ -}\Delta P \\ P^o=\text{ 17.5 mmHg - 0.3094 mmHg= 17.1906 mmHg} \end{gathered}[/tex]Then, the answer is that the vapor pressure of the urea solution in water is 17.1906 mmHg.
