JEE MAIN - Chemistry (2022 - 29th July Morning Shift - No. 16)
Resistance of a conductivity cell (cell constant $$129 \mathrm{~m}^{-1}$$) filled with $$74.5 \,\mathrm{ppm}$$ solution of $$\mathrm{KCl}$$ is $$100 \,\Omega$$ (labelled as solution 1). When the same cell is filled with $$\mathrm{KCl}$$ solution of $$149 \,\mathrm{ppm}$$, the resistance is $$50 \,\Omega$$ (labelled as solution 2). The ratio of molar conductivity of solution 1 and solution 2 is i.e. $$\frac{\wedge_{1}}{\wedge_{2}}=x \times 10^{-3}$$. The value of $$x$$ is __________. (Nearest integer)
Given, molar mass of $$\mathrm{KCl}$$ is $$74.5 \mathrm{~g} \mathrm{~mol}^{-1}$$.
Answer
1000
Explanation
$$\frac{l}{A}=129 \mathrm{~m}^{-1}$$
$$\mathrm{KCl}$$ solution $$1 \Rightarrow 74.5 \,\mathrm{ppm}, \mathrm{R}_{1}=100 \Omega$$
$$\mathrm{KCl}$$ solution $$2 \Rightarrow 149 \,\mathrm{ppm}, \mathrm{R}_{2}=50 \Omega$$
$$ \begin{aligned} &\text { Here, } \frac{p p m_{1}}{p p m_{2}}=\frac{M_{1}}{M_{2}}=\left(\frac{w_{1 / M_{0}}}{V} \times \frac{V}{w_{2 / M_{0}}}\right) \\ &\frac{\Lambda_{1}}{\Lambda_{2}}=\frac{k_{1} \times \frac{1000}{M_{1}}}{k_{2} \times \frac{1000}{M_{2}}} \\ &=\frac{k_{1}}{k_{2}} \times \frac{M_{1}}{M_{2}} \\ &=\frac{50}{100} \times 2 \\ &=\frac{\Lambda_{1}}{\Lambda_{2}}=1000 \times 10^{-3} \\ &=1000 \end{aligned} $$
$$\mathrm{KCl}$$ solution $$1 \Rightarrow 74.5 \,\mathrm{ppm}, \mathrm{R}_{1}=100 \Omega$$
$$\mathrm{KCl}$$ solution $$2 \Rightarrow 149 \,\mathrm{ppm}, \mathrm{R}_{2}=50 \Omega$$
$$ \begin{aligned} &\text { Here, } \frac{p p m_{1}}{p p m_{2}}=\frac{M_{1}}{M_{2}}=\left(\frac{w_{1 / M_{0}}}{V} \times \frac{V}{w_{2 / M_{0}}}\right) \\ &\frac{\Lambda_{1}}{\Lambda_{2}}=\frac{k_{1} \times \frac{1000}{M_{1}}}{k_{2} \times \frac{1000}{M_{2}}} \\ &=\frac{k_{1}}{k_{2}} \times \frac{M_{1}}{M_{2}} \\ &=\frac{50}{100} \times 2 \\ &=\frac{\Lambda_{1}}{\Lambda_{2}}=1000 \times 10^{-3} \\ &=1000 \end{aligned} $$
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