The equivalent conductance of an infinitely dilute solution of NH4Cl is 150 and the ionic conductance of OH- and Cl- ions are 198 and 76 respectively. If the equivalent conductance of a 0.01 N solution of NH4OH is 9.6. What will be its degree of dissociation?

Question

The equivalent conductance of an infinitely dilute solution of NH4Cl is 150 and the ionic conductance of OH- and Cl- ions are 198 and 76 respectively. If the equivalent conductance of a 0.01 N solution of NH4OH is 9.6. What will be its degree of dissociation? (A) 0.0353 (B) 0.0103 (C) 0.96 (D) 0.414

Tardigrade Admin · Accepted Answer

∧∞NH4OH =150-76 +198 =272 α=(9.6/272)=0.0353

Q. The equivalent conductance of an infinitely dilute solution of $N H_{4} Cl$ is $150$ and the ionic conductance of $O H^{-}$ and $C l^{-}$ ions are $198$ and $76$ respectively. If the equivalent conductance of a $0.01 N$ solution of $N H_{4} O H$ is $9.6$ . What will be its degree of dissociation?

$0.0353$

$0.0103$

$0.96$

$0.414$

$\land^{\infty} N H_{4} O H = 150 - 76 + 198 = 272$
$α = \frac{9.6}{272} = 0.0353$

Q. The equivalent conductance of an infinitely dilute solution of NH4​Cl is 150 and the ionic conductance of OH− and Cl− ions are 198 and 76 respectively. If the equivalent conductance of a 0.01N solution of NH4​OH is 9.6. What will be its degree of dissociation?

0.0353

0.0103

0.96

0.414

∧∞NH4​OH=150−76+198=272 α=2729.6​=0.0353

$0.0353$

$0.0103$

$0.96$

$0.414$

$\land^{\infty} N H_{4} O H = 150 - 76 + 198 = 272$
$α = \frac{9.6}{272} = 0.0353$