Question

Osmotic pressure of 40% (mass/vol.) urea solution is 1.64 atm and that of 3.42% (mass/vol.) canesugar is 2.46 atm. When equal volumes of the above two solutions are mixed, the osmotic pressure of the resulting solution is :

• A. $1.64$ atm
• B. $2.46$ atm
• C. $4.10$ atm
• D. $2.05$ atm

Answer 1

Answer: (D)

$2.05$ atm

The osmotic pressure of the resulting solution is 2.05 atm.
$\pi =\frac{{\pi }_1+{\pi }_2}{2}$, if equal volume are mixed; volume of solution become double.
$\pi$ = $1.64+2.46/2$
= $2.05atm$