Question

What will be the molar conductance $\text{'}\Lambda \text{'}$, if resistivity is $x$ for 0.1 $\text{N}{\text{H}}_{\text{2}}\text{S}{\text{O}}_{\text{4}}$?

• A. $\Lambda \text{=}\frac{x\times 1000}{0.1}$
• B. $\Lambda \text{=}\frac{2\times 1000}{x\times 0.1}$
• C. $\Lambda \text{=}\frac{x\times 1000}{0.5}$
• D. $\Lambda \text{=}\frac{0.5}{1000x}$

Answer 1

Answer: (B)

$\Lambda \text{=}\frac{2\times 1000}{x\times 0.1}$

Solution:- (B) $\Lambda =\frac{2\times 1000}{x\times 0.1}$

As we know that,

$N=n\times M$

Whereas,

$n=$ valence factor $=2(\because H_{2}S{O}_4\text{is a dibasic acid})$

$N=$ Normality $=0.1$

$M=$ Molarity $=?$

$\Rightarrow M=\frac{N}{n}=\frac{0.1}{2}$

Now, molar conductance is given as-

$\Lambda =\frac{1000\times k}{M}$

Whereas,

$\Lambda =$ Molar conductance $=?$

$k=$ Specific conductance $=\frac{1}{\text{specific resistance}}=\frac{1}{1}$

$M=$ Molarity $=\frac{0.1}{2}$

$\therefore \Lambda =\frac{1000\times \left(\frac{1}{x}\right)}{\left(\frac{0.1}{2}\right)}$

$\Rightarrow \Lambda =\frac{2\times 1000}{x\times 0.1}$

Hence the molar conductance is $\frac{2\times 1000}{x\times 0.1}$.