The osmotic pressure of equimolar solutions o | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Osmotic pressure $(\pi)$ is a colligative property,which depends on the van't Hoff factor $(i)$.$\pi = iCRT$.For equimolar solutions,$\pi \propto

Vedclass · Accepted Answer

$AlCl_3 > BaCl_2 > Urea$

Vedclass · Answer

(A) Osmotic pressure $(\pi)$ is a colligative property,which depends on the van't Hoff factor $(i)$.$\pi = iCRT$.For equimolar solutions,$\pi \propto i$.$Urea$ is a non-electrolyte,so $i = 1$.$BaCl_2$ dissociates as $BaCl_2 \rightarrow Ba^{2+} + 2Cl^-$,so $i = 3$.$AlCl_3$ dissociates as $AlCl_3 \rightarrow Al^{3+} + 3Cl^-$,so $i = 4$.Since the van't Hoff factor follows the order $AlCl_3 (4) > BaCl_2 (3) > Urea (1)$,the osmotic pressure follows the same order: $AlCl_3 > BaCl_2 > Urea$.

The osmotic pressure of equimolar solutions of urea,$BaCl_2$,and $AlCl_3$ will be in the order:

Similar Questions

$0.5 \ m$ aqueous solution of a weak acid $(HX)$ is $20 \%$ ionized. If $K_{f}$ of water is $1.86 \ K \ kg \ mol^{-1}$,the lowering in freezing point of the solution is: (in $K$)

$A$ $0.2 \, m$ aqueous solution of a weak acid $(HX)$ is $20 \% $ ionised. The freezing point of this solution is .............. $^o C$ (Given ${K_f} = 1.86 \, ^o C/m$ for water)

Which of the following equimolar aqueous solutions has the highest freezing point?

The experimental molecular weight of an electrolyte will always be less than its calculated value because the value of Van't Hoff factor $i$ is

Observe the following statements:
Statement-$I$: The boiling point of $0.1 \ M$ urea solution is less than that of $0.1 \ M$ $KCl$ solution.
Statement-$II$: Elevation of boiling point is inversely proportional to molar mass of solute.
The correct answer is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module