Chapter 10, Problem 37QAP

Interpretation Introduction

(a)

Interpretation:

Osmotic pressure of $\text{0}\text{.217}\text{M}$ solution of urea at $22\text{}°\text{C}$ is to be calculated.

Concept introduction:

Osmotic pressure is calculated by below formula.

$\pi =MRT$

(1)

Here, $\pi$ represents osmotic pressure, M represents molarity of solution, R and T represents gas constant and temperature.

Answer to Problem 37QAP

Osmotic pressure of $\text{0}\text{.217}\text{M}$ solution of urea at $22\text{}°\text{C}$ is $5.26\text{atm}$.

Explanation of Solution

Given, Temperature $=22°\text{C}$

Convert temperature from Celsius to Kelvin as follows:

$\begin{array}{c}T\text{=}\left(22+273\right)\text{K}\\ \text{=295}\text{K}\end{array}$

$\begin{array}{l}M=0.217\text{mol/L}\\ R=0.0821\text{L}\text{atm/mol}\text{K}\end{array}$

Substitute the values of T, M and R in equation (1).

$\begin{array}{c}\pi =MRT\\ =0.217\frac{\cancel{\text{mol}}}{\cancel{\text{L}}}\times \text{0}\text{.0821}\frac{\cancel{\text{L}}\text{atm}}{\cancel{\text{mol}}\cancel{\text{K}}}\times 295\cancel{\text{K}}\\ \text{=}5.26\text{atm}\end{array}$

Therefore, osmotic pressure of $\text{0}\text{.217}\text{M}$ urea at $22\text{}°\text{C}$ is $5.26\text{atm}$.

Interpretation Introduction

(b)

Interpretation:

The osmotic pressure of $\text{25}\text{.0}\text{g}$ of urea is to be calculated.

Concept introduction:

Expression of molarity of a solution is given below.

$\text{Molarity=}\frac{\text{moles}\text{of}\text{solute}}{\text{Volume}\text{of}\text{solution}\left(\text{L}\right)}$

Osmotic pressure of the solution is as follows:

$\pi =MRT$

Here, $\pi$ represents osmotic pressure, M represents molarity of solution, R and T represents gas constant and temperature.

Answer to Problem 37QAP

Osmotic pressure of $\text{25}\text{.0}\text{g}$ urea solution is $\text{14}\text{.70}\text{atm}$.

Explanation of Solution

Given:

mass of urea $=25.0\text{g}$ and

Volume of solution is $=685\text{mL}$.

$1\text{mol}{\left({\text{NH}}_{\text{2}}\right)}_{\text{2}}\text{CO}=60.05\text{g}{\left({\text{NH}}_{\text{2}}\right)}_{\text{2}}\text{CO}$

The conversion factor is $\frac{1\text{mol}{\left({\text{NH}}_{\text{2}}\right)}_{\text{2}}\text{CO}}{60.05\text{g}{\left({\text{NH}}_{\text{2}}\right)}_{\text{2}}\text{CO}}$

Use conversion factor to calculate moles of urea $\left(n\right)$ as follows:

$\begin{array}{c}n=25.0\text{g}\times \frac{1\text{mol}{\left({\text{NH}}_{\text{2}}\right)}_{\text{2}}\text{CO}}{60.05\text{g}{\left({\text{NH}}_{\text{2}}\right)}_{\text{2}}\text{CO}}\\ =0.416\text{mol}\end{array}$

Calculate molarity as follows:

$\text{Molarity=}\frac{\text{moles}\text{of}\text{solute}}{\text{Volume}\text{of}\text{solution}\left(\text{L}\right)}$.. ..... (2)

$\begin{array}{c}\text{Given, Volume}\text{=}68\text{5}\text{mL}\\ \text{1000}\text{mL =}\text{1}\text{L}\\ 685\text{mL}=\frac{685}{1000}\text{L}\\ \text{=0}\text{.685}\text{L}\end{array}$

Moles of solute ( n) $=0.416\text{mol}$

Put the above values in equation (2).

$\begin{array}{c}\text{Molarity=}\frac{\text{moles}\text{of}\text{solute}}{\text{Volume}\text{of}\text{solution}\left(\text{L}\right)}\\ \text{=}\frac{0.416\text{mol}}{0.685\text{L}}\\ =0.607\text{mol/L}\end{array}$

Osmotic pressure of the solution as follows:

$\pi =MRT$

(3)

Here, $\pi$ represents osmotic pressure, M represents molarity of solution, R and T represents gas constant and temperature Calculate osmotic pressure of urea as follows:

Given, Temperature $=22°\text{C}$

Convert temperature from Celsius to Kelvin as follows:

$\begin{array}{c}T\text{=}\left(22+273\right)\text{K}\\ \text{=295}\text{K}\end{array}$

$\begin{array}{l}M=0.607\text{mol/L}\\ R=0.0821\text{L}\text{atm/mol}\text{K}\end{array}$

Substitute the values of T, M and R in equation (3).

$\begin{array}{c}\pi =MRT\\ =0.607\frac{\cancel{\text{mol}}}{\cancel{\text{L}}}\times \text{0}\text{.0821}\frac{\cancel{\text{L}}\text{atm}}{\cancel{\text{mol}}\cancel{\text{K}}}\times 295\cancel{\text{K}}\\ \text{=14}\text{.70}\text{atm}\end{array}$

Therefore, osmotic pressure of $\text{25}\text{.0}\text{g}$ urea solution is $\text{14}\text{.70}\text{atm}$.

Interpretation Introduction

(c)

Interpretation:

The osmotic pressure of $15.0\%$ urea by mass solution is to be calculated.

Concept introduction:

Relationship between density, volume and mass is given below.

$\text{Density=}\frac{\text{Mass}}{\text{Volume}}$

Answer to Problem 37QAP

Osmotic pressure of $15.0\%$ solution of urea at temperature $22\text{}°\text{C}$ is $\text{67}.8\text{atm}$.

Explanation of Solution

Given:

15.0 % urea by mass.

Density of solution = 1.12 g/mL.

Molar mass of urea $=60.05\text{g}$

Given, mass $=15\text{g}$

Therefore,

$1\text{mol}{\left({\text{NH}}_{\text{2}}\right)}_{\text{2}}\text{CO}=60.05\text{g}{\left({\text{NH}}_{\text{2}}\right)}_{\text{2}}\text{CO}$

The conversion factor is $\frac{1\text{mol}{\left({\text{NH}}_{\text{2}}\right)}_{\text{2}}\text{CO}}{60.05\text{g}{\left({\text{NH}}_{\text{2}}\right)}_{\text{2}}\text{CO}}$

Use conversion factor to calculate moles of urea $\left(n\right)$ as follows:

$\begin{array}{c}n=15\text{g}{\left({\text{NH}}_{\text{2}}\right)}_{\text{2}}\text{CO}\times \frac{1\text{mol}{\left({\text{NH}}_{\text{2}}\right)}_{\text{2}}\text{CO}}{60.05\text{g}{\left({\text{NH}}_{\text{2}}\right)}_{\text{2}}\text{CO}}\\ =0.250\text{mol}{\left({\text{NH}}_{\text{2}}\right)}_{\text{2}}\text{CO}\end{array}$

$\begin{array}{l}\text{Given},\\ \text{Density of the solution}=1.12\text{g/mL}\\ \text{Mass of solution}=100\text{g}\end{array}$

The relationship between density, volume and mass is given below.

$\text{Density=}\frac{\text{Mass}}{\text{Volume}}$.. .... (4)

Put above values in (4).

$\begin{array}{c}\text{Volume=}\frac{\text{Mass}}{\text{density}}\\ =\frac{100\text{g}}{1.12\text{g/mL}}\\ =89.3\text{mL}\end{array}$

$\begin{array}{c}1000\text{mL=1}\text{L}\\ \text{89}\text{.3}\text{mL}=\frac{89.3}{1000}\text{L}\\ \text{=0}\text{.0893}\text{L}\end{array}$

Therefore, volume of the solution is $\text{0}\text{.0893}\text{L}$.

Calculate molarity as follows:

$\text{Molarity=}\frac{\text{moles}\text{of}\text{solute}}{\text{Volume}\text{of}\text{solution}\left(\text{L}\right)}$.. ..... (5)

$\text{Volume}\text{=}0.0893\text{L}$

Moles of solute ( n) $=0.250\text{mol}$

Put the above values in equation (5).

$\begin{array}{c}\text{Molarity=}\frac{\text{moles}\text{of}\text{solute}}{\text{Volume}\text{of}\text{solution}\left(\text{L}\right)}\\ \text{=}\frac{0.250\text{mol}}{0.0893\text{L}}\\ =2.80\text{mol/L}\end{array}$

Calculate osmotic pressure of the solution as follows:

Given, Temperature $=22°\text{C}$

Convert temperature from Celsius to Kelvin as follows:

$\begin{array}{c}T\text{=}\left(22+273\right)\text{K}\\ \text{=295}\text{K}\end{array}$

$\begin{array}{l}M=2.80\text{mol/L}\\ R=0.0821\text{L}\text{atm/mol}\text{K}\end{array}$

Substitute the values of T, M and R in below expression.

$\begin{array}{c}\pi =MRT\\ =2.80\frac{\cancel{\text{mol}}}{\cancel{\text{L}}}\times \text{0}\text{.0821}\frac{\cancel{\text{L}}\text{atm}}{\cancel{\text{mol}}\cancel{\text{K}}}\times 295\cancel{\text{K}}\\ \text{=67}.8\text{atm}\end{array}$

Therefore, osmotic pressure of $15.0\%$ solution of urea at temperature $22\text{}°\text{C}$ is $\text{67}.8\text{atm}$.