Question

0.004M Na2So4 is isotonic with 0.01M glucose. Degree of dissociation of Na2So4 is

Answer 1

The reaction is given as,

i$c_{1}$RT=$c_{2}$RT

i$c_{1}$=$c_{2}$

(1+2$\alpha$)0.004=0.01

$\alpha$=0.75

hence degree of dissociation is 75%

Answer 2

Answer: 0.006

Explanation: Isotonic solutions are those solutions which have the same osmotic pressure.  If osmotic pressures are equal at the same temperature, concentrations must also be equal.

$\pi =CRT$ for non electrolytes such as glucose

$\pi =iCRT$ for electrolytes such as $Na_2SO_4$

$\pi$ = osmotic pressure

C= concentration

R= solution constant

T= temperature

i= vant hoff factor

Thus: $C_{urea}=iC_{Na_2SO_4}$ where concentration is in molarity.

$0.01M=i\times 0.004M$

$i=2.5$

$i=\frac{\text {observed colligative property}}{\text {Calculated Colligative property}}$

$Na_2SO_4\rightarrow 2Na^++SO_4^{2-}$

  0.004             0             0

$0.004-\alpha$        $\alpha$          $\alpha$  

Total moles after dissociation =$0.004-\alpha+\alpha+\alpha=0.004+\alpha$  

thus $i=\frac{0.004+\alpha}{0.004}$

$2.5=\frac{0.004+\alpha}{0.004}$

$\alpha=0.006$.