Question

A compound is found to have a molar mass of 598 g/mol. If 35.8 mg of the compound dissolved in enough water to make 175 ml of solution at 25 degrees C what is the osmotic pressure of the resulting solution ?

Answer 1

Concepts and reason

This problem is based on the concept osmotic pressure.

Osmotic pressure can be calculated with the help of concentration and temperature.

Fundamentals

The minimum pressure which is applied in solution side in order to prevent the inward flow of pure solvent is known as osmotic pressure$\left( \Pi \right)$. Osmotic pressure depends on the concentration as well as temperature. The formula to calculate osmotic pressure is as follows:

$\Pi = CRT$ …… (1)
‎Here,$C$is concentration, $R$is gas constant and $T$is temperature.

The relation between $C$ and volume$\left( V \right)$is as follows:

$C = \frac{n}{V}$ …… (2)

Here,$n$is the number of moles.

Also, number of moles is terms of molar mass$\left( M \right)$is as follows:

$n = \frac{m}{M}$ …… (3)

Here,$m$is mass.

The mass and molar mass of compound are $35.8{\rm{ mg}}$ and ${\rm{598 g/mol}}$respectively. The mass is in milligrams, convert it into grams. The relation between grams $\left( {\rm{g}} \right)$and milligrams$\left( {{\rm{mg}}} \right)$ is as follows:

$1{\rm{ mg}} = {10^{ - 3}}{\rm{ g}}$

The mass of the compound in grams is as follows:

$\begin{array}{c}\\m = 35.8{\rm{ mg}}\left( {\frac{{{{10}^{ - 3}}{\rm{ g}}}}{{1{\rm{ mg}}}}} \right)\\\\ = 35.8{\rm{ }} \times {10^{ - 3}}{\rm{ g}}\\\end{array}$

Substitute $35.8{\rm{ }} \times {10^{ - 3}}{\rm{ g}}$for $m$and ${\rm{598 g/mol}}$ for $M$in equation (3) as follows;

$\begin{array}{c}\\n = \frac{{35.8{\rm{ }} \times {{10}^{ - 3}}{\rm{ g}}}}{{{\rm{598 g/mol}}}}\\\\ = 0.0599 \times {10^{ - 3}}{\rm{ mol}}\\\end{array}$

The volume is in milliliters. Convert it into liters. The relation between milliliters$\left( {{\rm{mL}}} \right)$and liters$\left( {\rm{L}} \right)$is as follows:

${\rm{1 mL}} = {10^{ - 3}}{\rm{ L}}$

So, the volume in liters is as follows:

$\begin{array}{c}\\V = 175{\rm{ mL}}\left( {\frac{{{{10}^{ - 3}}{\rm{ L}}}}{{{\rm{1 mL}}}}} \right)\\\\ = 175 \times {10^{ - 3}}{\rm{ L}}\\\end{array}$

Substitute $0.0599 \times {10^{ - 3}}{\rm{ mol}}$ for $n$and $175 \times {10^{ - 3}}{\rm{ L}}$ for $V$in equation (2) as follows:

$\begin{array}{c}\\C = \frac{{0.0599 \times {{10}^{ - 3}}{\rm{ mol}}}}{{175 \times {{10}^{ - 3}}{\rm{ L}}}}\\\\ = 3.42 \times {10^{ - 4}}{\rm{ mol/L}}\\\end{array}$

Temperature is in Celsius. Convert it into Kelvin as follows:

$\begin{array}{c}\\T = \left( {25 + 273.15} \right){\rm{ K}}\\\\ = {\rm{298}}{\rm{.15 K}}\\\end{array}$

Substitute $3.42 \times {10^{ - 4}}{\rm{ mol/L}}$ for $C$,$0.08206{\rm{ L atm mo}}{{\rm{l}}^{ - 1}}{\rm{ }}{{\rm{K}}^{ - 1}}$ for $R$and ${\rm{298}}{\rm{.15 K}}$for $T$ in equation (1) as follows:

$\begin{array}{c}\\\Pi = \left( {3.42 \times {{10}^{ - 4}}{\rm{ mol/L}}} \right)\left( {0.08206{\rm{ L atm mo}}{{\rm{l}}^{ - 1}}{\rm{ }}{{\rm{K}}^{ - 1}}} \right)\left( {{\rm{298}}{\rm{.15 K}}} \right)\\\\ = 83.6 \times {10^{ - 4}}{\rm{ atm}}\\\end{array}$

Unit of osmotic pressure is ${\rm{torr}}$. Relation between ${\rm{atm}}$and ${\rm{torr}}$ is as follows:

${\rm{1 atm}} = 760{\rm{ torr}}$

Convert pressure into ${\rm{torr}}$as follows:

$\begin{array}{c}\\\Pi = 83.6 \times {10^{ - 4}}{\rm{ atm}}\left( {\frac{{760{\rm{ torr}}}}{{{\rm{1 atm}}}}} \right)\\\\ = 6.36{\rm{ torr}}\\\end{array}$

Ans:

The osmotic pressure of the resulting solution is $6.36{\rm{ torr}}$.