Question

What is the osmotic pressure of 0.9% m/v solution of NaCl at 25°C?

Answer 1

Osmotic pressure $= i \times C \times R \times T$

where, C= concentration of solute(in terms of Molarity)

$R = 0.082$

$i = 1$

0.9% $NaCl$ solution means 0.9g is present in 100mL of solution.

Moles of $NaCl$ $= \dfrac{0.9}{58} = 0.015$

Concentration of $NaCl = \dfrac{0.015}{100} \times 1000 = 0.10$

Hence, Osmotic pressure $= 1 \times 0.10 \times 0.082 \times 300$

$= 2.46atm$