6. ⇒ (MHT CET 2021 21th September Evening Shift )

What will be the concentration of solution of electrolyte if it's molar conductivity and conductivity are respectively 230 $Ω^{- 1}$ cm $^{2}$ mol $^{- 1}$ and 0.0115 $Ω^{- 1}$ cm $^{- 1}$ at 298 K?

A. 0.04 M

B. 0.03 M

C. 0.01 M

D. 0.05 M

Correct Option is (D)

$\begin{aligned} \land = & 230 Ω^{- 1} {cm}^{2} {mol}^{- 1}, k = 0.0115 Ω^{- 1} {cm}^{- 1} \\ \land & = \frac{1000 k}{c} \\ ∴ c & = \frac{1000 k}{\hat{N}} = \frac{1000 {cm}^{3} L^{- 1} \times 0.0115 Ω^{- 1} {cm}^{- 1}}{230 Ω^{- 1} {cm}^{2} {mol}^{- 1}} \\ = 0.05 mol L^{- 1} \end{aligned}$

7. ⇒ (MHT CET 2021 21th September Morning Shift )

Which among the following statements is true for conductivity?

A. It is inversely proportional to resistivity.

B. It is inversely proportional to molar conductivity.

C. It is directly proportional to resistivity.

D. It is directly proportional to resistance.

Correct Option is (A)

Conductivity is inversely proportional to resistivity i.e. $k = \frac{1}{ρ}$

8. ⇒ (MHT CET 2021 21th September Morning Shift )

The molar conductivity of $0.4 M KCl$ solution is $2.5 \times 10^{5} Ω^{- 1} {cm}^{2} {mol}^{- 1}$ . What is the resistivity of solution?

A. $2.1 \times 10^{2}$

B. $2.5 \times 10^{2}$

C. $1 \times 10^{- 2}$

D. $2.8 \times 10^{- 2}$

Correct Option is (C)

$\begin{aligned} c = 0.4 M, \land = 2.5 \times 10^{5} Ω^{- 1} {cm}^{2} {mol}^{- 1}, ρ = ? \end{aligned}$

$\begin{aligned} (i) \land = \frac{1000 k}{C} ∴ k = \frac{\land C}{1000} ∴ k = \frac{2.5 \times 10^{5} \times 0.4}{1000} = 100 \end{aligned}$

(ii) $k = \frac{1}{ρ} ∴ ρ = \frac{1}{k} = \frac{1}{100} = 1 \times 10^{- 2}$

9. ⇒ (MHT CET 2021 20th September Evening Shift )

What is the molar conductivity of $0.05 M$ solution of sodium hydroxide, if it's conductivity is $0.0118 S {cm}^{- 1}$ at $298 K$ ?

A. $236 S {cm}^{2} {mol}^{- 1}$

B. $423 S {cm}^{2} {mol}^{- 1}$

C. $354 S {cm}^{2} {mol}^{- 1}$

D. $590 S {cm}^{2} {mol}^{- 1}$

Correct Option is (A)

$\begin{aligned} c = & 0.05 M, k = 0.0118 S {cm}^{- 1} \\ \land & = \frac{1000 k}{c} \\ = \frac{1000 {cm}^{3} L^{- 1} \times 0.0118 {S cm}^{- 1}}{0.05 mol L^{- 1}} \\ = 236 S {cm}^{2} {mol}^{- 1} \end{aligned}$

10. ⇒ (MHT CET 2021 20th September Morning Shift )

The molar conductivity of 0.1 M BaCl $_{2}$ solution is 106 $Ω^{- 1}$ cm $^{2}$ mol $^{- 1}$ at 25 $^{\circ}$ C. What is it's conductivity?

A. 1.06 $\times$ 10 $^{- 2}$ $Ω^{- 1}$ cm $^{- 1}$

B. 5.3 $\times$ 10 $^{- 3}$ $Ω^{- 1}$ cm $^{- 1}$

C. 3.66 $\times$ 10 $^{- 3}$ $Ω^{- 1}$ cm $^{- 1}$

D. 2.6 $\times$ 10 $^{- 2}$ $Ω^{- 1}$ cm $^{- 1}$

Correct Option is (A)

$\land = \frac{1000 k}{c} ∴ k = \frac{\land c}{1000}$

Now, $\land = 106 Ω^{- 1} {cm}^{2} {mol}^{- 1}, c = 0.1 mol L^{- 1}$

$\begin{aligned} ∴ k & = \frac{106 Ω^{- 1} {cm}^{2} {mol}^{- 1} \times 0.1 mol L^{- 1}}{1000 {cm}^{3} L^{- 1}} \\ ∴ k & = 1.06 \times 10^{- 2} Ω^{- 1} {cm}^{- 1} \end{aligned}$

⇒Electrochemistry