Question

When 4.33g of a nonelectrolyte solute is dissolved in water to make 205 mL of solution at 21 C, the solution exerts an osmotic pressure of 891 torr.

What is the molar concentration of the solution?

How many moles of solute are in the solution?

What is the molar mass of the solute?

Answer 1

Concepts and reason

The concepts used to solve this problem are osmotic pressure and molarity.

The osmotic pressure is a colligative property as it is directly proportional to the concentration of solute. The number of moles of solute can be calculated by multiplying the molarity and volume of solution in liters. The molar mass of solute is obtained by dividing the mass of solute by the number of moles of solute.

Fundamentals

The osmotic pressure $\left( \prod \right)$ is calculated as:

$\prod = {C_{{\rm{solute}}}}RT$

Here, ${C_{{\rm{solute}}}}$ is the molar concentration of solute, $R$ is the universal gas constant which is equal to $0.0821{\rm{ L atm mo}}{{\rm{l}}^{ - 1}}{{\rm{K}}^{ - 1}}$ and $T$ is the temperature in Kelvin.

‎The number of moles of solute $\left( n \right)$ is calculated as:

$n = M \times V$

Here, $M$ is the molarity in ${\rm{mol/L}}$ and $V$ is the volume of solution in ${\rm{L}}$ .

The molar mass $\left( M \right)$ of solute is calculated as:

$M = \frac{m}{n}$

Here, $m$ is the mass of solute and $n$ is the number of moles of solute.

(1)

The osmotic pressure $\left( \prod \right)$ of the solution is $891{\rm{ torr}}$

Since,

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

Therefore,

$\begin{array}{c}\\\prod = 891{\rm{ torr}} \times \frac{{1{\rm{ atm}}}}{{760{\rm{ torr}}}}\\\\ = 1.17{\rm{ atm}}\\\end{array}$

The temperature $\left( T \right)$ is $21^\circ {\rm{C}}$ .

Since,

$T\left( {\rm{K}} \right) = T\left( {{\rm{^\circ C}}} \right) + 273.15$

Therefore,

$\begin{array}{c}\\T\left( {\rm{K}} \right) = 21^\circ {\rm{C}} + 273.15\\\\ = 294.15{\rm{ K}}\\\end{array}$

The expression of osmotic pressure $\left( \prod \right)$ is:

$\prod = CRT$

On rearranging the above expression,

$C = \frac{\prod }{{RT}}$

Substitute $1.17{\rm{ atm}}$ for $\prod$ , $0.0821{\rm{ L atm mo}}{{\rm{l}}^{ - 1}}{{\rm{K}}^{ - 1}}$ for $R$ and $294.15{\rm{ K}}$ for $T$ in the above expression.

$\begin{array}{c}\\C = \frac{{1.17{\rm{ atm}}}}{{\left( {0.0821{\rm{ L atm mo}}{{\rm{l}}^{ - 1}}{{\rm{K}}^{ - 1}}} \right) \times \left( {294.15{\rm{ K}}} \right)}}\\\\ = 0.0485{\rm{ mol/L}}\\\\ = 0.0485{\rm{ M}}\\\end{array}$

The molar concentration of the solution is $0.0485{\rm{ M}}$ .

(2)

The volume $\left( V \right)$ of the solution is $205{\rm{ mL}}$ .

Since,

$1{\rm{ mL}} = {\rm{0}}{\rm{.001 L}}$

Therefore,

$205{\rm{ mL}} = {\rm{0}}{\rm{.205 L}}$

Thus, the volume of solution in liters is ${\rm{0}}{\rm{.205 L}}$ .

The number of moles of solute $\left( n \right)$ is calculated as:

$n = M \times V$

Substitute $0.0485{\rm{ mol/L}}$ for $M$ and ${\rm{0}}{\rm{.205 L}}$ for $V$ .

$\begin{array}{c}\\n = 0.0485{\rm{ mol/L}} \times {\rm{0}}{\rm{.205 L}}\\\\ = 0.00994{\rm{ mol}}\\\end{array}$

Thus, the number of moles of nonelectrolyte solute is $0.00994{\rm{ mol}}$ .

(3)

The mass of nonelectrolyte solute $\left( m \right)$ dissolved in water $4.33{\rm{ g}}$ .

The molar mass $\left( M \right)$ of solute is calculated as:

$M = \frac{m}{n}$

Substitute $4.33{\rm{ g}}$ for $m$ and $0.00994{\rm{ mol}}$ for $n$ .

$\begin{array}{c}\\M = \frac{{4.33{\rm{ g}}}}{{0.00994{\rm{ mol}}}}\\\\ = 435.6{\rm{ g/mol}}\\\\ \approx {\rm{436 g/mol}}\\\end{array}$

Therefore, the molar mass of solute is ${\rm{436 g/mol}}$ .

Ans:

The molar concentration of the solution is $0.0485{\rm{ M}}$ .