Solved: CHEMISTRY Question for NEET

NEET CHEMISTRYMedium

Limiting molar conductivities, for the given solutions, are: $\lambda^0_m(\text{H}_2\text{SO}_4) = x \text{ S cm}^2 \text{ mol}^{-1}$ $\lambda^0_m(\text{K}_2\text{SO}_4) = y \text{ S cm}^2 \text{ mol}^{-1}$ $\lambda^0_m(\text{CH}_3\text{COOK}) = z \text{ S cm}^2 \text{ mol}^{-1}$ From the data given above, it can be concluded that $\lambda^0_m$ in $(\text{S cm}^2 \text{ mol}^{-1})$ for $\text{CH}_3\text{COOH}$ will be:

$x - y + 2z$

$x + y + z$

$x - y + z$

$\frac{(x - y)}{2} + z$

Step-by-Step Solution

According to Kohlrausch's law of independent migration of ions : $\lambda^0_m(\text{H}_2\text{SO}_4) = 2\lambda^0_m(\text{H}^+) + \lambda^0_m(\text{SO}_4^{2-}) = x$ --- (i) $\lambda^0_m(\text{K}_2\text{SO}_4) = 2\lambda^0_m(\text{K}^+) + \lambda^0_m(\text{SO}_4^{2-}) = y$ --- (ii) $\lambda^0_m(\text{CH}_3\text{COOK}) = \lambda^0_m(\text{CH}_3\text{COO}^-) + \lambda^0_m(\text{K}^+) = z$ --- (iii)

We need to find the limiting molar conductivity of acetic acid ( $\text{CH}_3\text{COOH}$ ): $\lambda^0_m(\text{CH}_3\text{COOH}) = \lambda^0_m(\text{CH}_3\text{COO}^-) + \lambda^0_m(\text{H}^+)$

Subtracting equation (ii) from (i): $2\lambda^0_m(\text{H}^+) - 2\lambda^0_m(\text{K}^+) = x - y$ $\lambda^0_m(\text{H}^+) - \lambda^0_m(\text{K}^+) = \frac{x - y}{2}$ --- (iv)

Now, adding equation (iii) and (iv): $\lambda^0_m(\text{CH}_3\text{COO}^-) + \lambda^0_m(\text{K}^+) + \lambda^0_m(\text{H}^+) - \lambda^0_m(\text{K}^+) = z + \frac{x - y}{2}$ $\lambda^0_m(\text{CH}_3\text{COO}^-) + \lambda^0_m(\text{H}^+) = \frac{(x - y)}{2} + z$ $\lambda^0_m(\text{CH}_3\text{COOH}) = \frac{(x - y)}{2} + z$

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