"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 $74.5 \mathrm{~g} \mathrm{~mol}^{-1}$."

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"$\frac{l}{A}=129 \mathrm{~m}^{-1}$

$\mathrm{KCl}$ solution $1 \Rightarrow 74.5 \,\mathrm{ppm}, \mathrm{R}_{1}=100 \Omega$

$\mathrm{KCl}$ solution $2 \Rightarrow 149 \,\mathrm{ppm}, \mathrm{R}_{2}=50 \Omega$

$$ \begin{aligned} &\text { Here, } \frac{p p m_{1}}{p p m_{2}}=\frac{M_{1}}{M_{2}}=\left(\frac{w_{1 / M_{0}}}{V} \times \frac{V}{w_{2 / M_{0}}}\right) \\ &\frac{\Lambda_{1}}{\Lambda_{2}}=\frac{k_{1} \times \frac{1000}{M_{1}}}{k_{2} \times \frac{1000}{M_{2}}} \\ &=\frac{k_{1}}{k_{2}} \times \frac{M_{1}}{M_{2}} \\ &=\frac{50}{100} \times 2 \\ &=\frac{\Lambda_{1}}{\Lambda_{2}}=1000 \times 10^{-3} \\ &=1000 \end{aligned} $$"

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