Question

The osmotic pressure of a 0.9 weight percent solution of sodium chloride (NaCl) is equal to the osmotic pressure of a 5 weight percent solution of non-dissociating molecule X. What is the molecular weight of molecule X

Answer 1

Answer: The molar mass of molecule X is 162.5 g/mol

Explanation:

We are given:

0.9 % of NaCl

This means that 0.9 grams of NaCl is present in 100 grams of solution

5 % of molecule X

This means that 5 grams of molecule X is present in 100 grams of solution

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

$\pi=iMRT$

Or,

$\pi=i\times \frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}\times RT$

where,

i = Van't hoff factor

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature of the solution

According to the question:

Osmotic pressure of both the solutions are same.

So, the equation becomes:

$i_{NaCl}\times \frac{m_{NaCl}}{M_{NaCl}\times V}\times RT=i_X\times \frac{m_X}{M_X\times V}\times RT$

where,

$i_{NaCl}=2\\m_{NaCl}=0.9g\\M_{NaCl}=58.5g/mol\\i_X=1\text{ (for non-electrolytes)}\\m_X=5g\\M_X=?g/mol$

Putting values in above equation, we get:

$2\times \frac{0.9}{58.5\times V}\times RT=1\times \frac{5}{M_X\times V}\times RT\\\\M_X=\frac{1\times 5\times 58.5}{0.9\times 2}=162.5g/mol$

Hence, the molar mass of molecule X is 162.5 g/mol