Chapter 12, Problem 12.54QP

Interpretation Introduction

Interpretation:

The grams of urea required to prepare solution in $450g$ of water with given temperature and vapor pressure conditions should be calculated.

Concept introduction:

Raoult’s law: Raoult’s law states that solvent partial pressure over a solution is equal to vapor pressure of pure solvent multiplied with mole fraction of the solvent.

${\text{P}}_{\text{1}}{\text{= χ}}_{\text{1}}{\text{P}}^{\text{°}}{}_{\text{1}}$

Where, ${\text{P}}_{\text{1}}=$ Partial pressure of solvent

${\text{P}}_{\text{1}}^{\circ }=$ Partial pressure of its pure solvent

${\text{χ}}_{\text{1}}=$ Mole fraction of the solvent

Moles: One mole is equivalent to the mass of the substance consists same number of units equal to the atoms present in $\text{12}\text{g}$ of ${}^{\text{12}}\text{C}$.

From given mass of substance moles could be calculated by using the following formula,

$\text{Moles}\text{of}\text{substance }\text{= }\frac{\text{Given}\text{mass}\text{of}\text{substance}}{\text{Molecular}\text{mass}}$

Mass: It is the quantitative measure of a substance. The amount of matter present in substance is expressed as mass. The $S.I.$ unit of mass is $kg$.

Density is calculated by mass of the substance divided by its volume.

$\text{density =}\frac{\text{mass}}{\text{volume}}$

Answer to Problem 12.54QP

The grams of urea required to prepare solution is $\text{128}\text{g}$.

Explanation of Solution

Given data: Mass of ${\text{H}}_{\text{2}}\text{O}=450g$

Vapor pressure $=2.50mmHg$

Temperature ${\text{= 30}}^{\text{o}}\text{C}$

Vapor pressure of ${\text{H}}_{\text{2}}\text{O}\text{at}{\text{30}}^{\text{o}}\text{C}$$=31.8mmHg$

Calculate mole fraction of urea:

Substitute the solvent vapor pressure value and vapor pressure value of pure solvent into Raoutl’s Law equation to calculate mole fraction.

$\begin{array}{c}{\text{P}}_{\text{1}}{\text{ = χ}}_{\text{1}}{\text{P}}^{\text{°}}{}_{\text{1}}\\ \text{2}{\text{.50 mmHg = χ}}_{\text{urea}}\left(\text{31}\text{.8 mmHg}\right)\\ {\text{χ}}_{\text{urea}}=0.0786\end{array}$

Calculate moles of water:

$\begin{array}{c}\text{Moles}\text{of}\text{substance }\text{= }\frac{\text{Given}\text{mass}\text{of}\text{substance}}{\text{Molecular}\text{mass}}\\ H_{2}OMoles=\frac{\text{450}\cancel{\text{g}}}{\text{18}\text{.02}\cancel{\text{g}}\text{/mol}}=25.0mol\end{array}$

Calculate moles of urea:

Substitute the mole fraction of urea and the water moles into mole fraction equation to calculate the moles of urea.

$\begin{array}{c}{\text{X}}_{\text{urea}}\text{= }\frac{{\text{n}}_{\text{urea}}}{{\text{n}}_{\text{water}}{\text{+n}}_{\text{urea}}}\\ 0.0786=\frac{{\text{n}}_{\text{urea}}}{\text{25}{\text{.0+ n}}_{\text{urea}}}\\ 0.0786\left(\text{25}{\text{.0+ n}}_{\text{urea}}\right)={\text{n}}_{\text{urea}}\\ 1.965+0.0786{\text{n}}_{\text{urea}}={\text{n}}_{\text{urea}}\end{array}$

$\begin{array}{c}1.965={\text{n}}_{\text{urea}}-0.0786{\text{n}}_{\text{urea}}\\ 1.965={\text{n}}_{\text{urea}}\left(1-0.0786\right)\\ {\text{n}}_{\text{urea}}=\frac{1.965}{\left(1-0.0786\right)}\\ {\text{n}}_{\text{urea}}=2.13mol\end{array}$

Convert moles of urea into mass:

$\begin{array}{c}\text{Given}\text{mass}\text{of}\text{substance}=\text{Moles}\text{of}\text{substance}\times \text{Molecular}\text{mass}\\ =2.13\cancel{mol}\times 60.06g/\cancel{mol}\\ =128g\end{array}$

Therefore, the mass of urea is $128g$.

Conclusion

The grams of urea required to prepare solution in $450g$ of water with given temperature and vapor pressure conditions was calculated.