Question

$0.585$% $NaCl$ solution at ${27}^{\circ }$C has osmotic pressure of:

• A. $2.49$ atm
• B. $4.92$ atm
• C. $1.2$ atm
• D. $3.8$ atm

Answer 1

Answer: (B)

$4.92$ atm

Given that

$T=27+273=300K$

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

where,

$i=2$ ($\because$ van't Hoff factor for electrolyte $=2$ in this case)

$0.585\%$ NaCl $=0.585$g NaCl (Given)

Mole of NaCl $=\frac{0.585}{58}=0.01mol$ (mol. weight$=58$ of NaCl)

$\therefore C=\frac{0.01}{100}\times 1000=0.1$

$\therefore$ Osmotic pressure $=2\times 0.1\times 0.082\times 300$

$=4.92$ atm.