Question

Osmotic pressure of a 4.44% solution of anhydrous CaCl2 was found to be 16.42 atm at 27°C. What is the degree of dissociation of CaCl2 ?

Answer 1

α(degree of dissociation )= i-1/n-1 Where i=van hoff factor n= no of moles after dissociation on calculating i=1.66 n=2 α=33.33%

Answer 2

Answer:

the degree of dissociation of CaCl2 is $0.24$

Explanation:

Given that

$4.44\%$solution

$\Rightarrow 5 \mathrm{~g}$in $100 \mathrm{ml} solum$

$\begin{aligned} \omega_{B} &=s g \\ M_{B} &=16.42 \mathrm{~g} / \mathrm{mol} \\ T &=0^{\circ} \mathrm{K}=27 \mathrm{~K} \end{aligned}$

N.w

$\begin{aligned}\Pi &=i C R T=i \frac{\omega_{B}}{M_{0} \times V} R T \\15 &=i \times \frac{5}{111 \times 0.2 \times 0.082 \times 2.73} \\j &=\frac{15 \times 111 \times 0.2}{5 \times 0.082 \times 273}\end{aligned}$

$\approx 1.49$

$\begin{aligned} N_{0 \omega} & \\ & \mathrm{CaCl}_{2} \longrightarrow \end{aligned}$

Nov, $\quad i=1+2 \alpha$

$=0.24$