0.004 M Na_2SO_4 solution is isotonic with 0 | vedclass

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

Question

(C) Since the solutions are isotonic,their osmotic pressures are equal: $\pi_{Na_2SO_4} = \pi_{Glucose}$.For glucose (non-electrolyte),$\pi = C 	imes

Vedclass · Accepted Answer

$75$

Vedclass · Answer

(C) Since the solutions are isotonic,their osmotic pressures are equal: $\pi_{Na_2SO_4} = \pi_{Glucose}$.For glucose (non-electrolyte),$\pi = C 	imes R 	imes T = 0.01 	imes R 	imes T$.For $Na_2SO_4$,$\pi = i 	imes C 	imes R 	imes T = i 	imes 0.004 	imes R 	imes T$.Equating the two: $i 	imes 0.004 = 0.01$,so $i = 2.5$.For $Na_2SO_4 ightarrow 2 Na^{+} + SO_4^{2-}$,the van't Hoff factor $i = 1 + (n-1)\alpha$,where $n = 3$.$2.5 = 1 + (3-1)\alpha = 1 + 2\alpha$.$2\alpha = 1.5$,which gives $\alpha = 0.75$.Therefore,the degree of dissociation is $75 \%$.

$0.004 \ M$ $Na_2SO_4$ solution is isotonic with $0.01 \ M$ glucose solution,then the degree of dissociation of $Na_2SO_4$ is ........... $\%$.

Similar Questions

If sodium sulfate $(Na_2SO_4)$ undergoes complete dissociation into its constituent ions in an aqueous solution,what will be the depression in freezing point $(\Delta T_f)$ when $0.01 \ mol$ of sodium sulfate is dissolved in $1 \ kg$ of water (in $K$)? (Given: $K_f = 1.86 \ K \ kg \ mol^{-1}$)

Which solution has the maximum boiling point?

Van't Hoff factor of $BaCl_2$ in its aqueous solution will be: ($BaCl_2$ is $60 \%$ ionized in the solution)

Calculate the freezing point (in ${}^{\circ}C$) of a solution obtained by dissolving $0.1 \ g$ of potassium ferricyanide (molecular weight $= 329$) in $100 \ g$ of water. Given that $K_f$ for water is $1.86 \ K \ kg \ mol^{-1}$.

$A$ $0.1 \ m$ aqueous solution of a weak acid $HX$ is $30\%$ ionized. If $K_f$ for water is $1.86 \ ^{\circ}C/m,$ the freezing point of the solution will be ......... $^{\circ}C$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module