Weight Percentage of Salt from Conductance and Molar Conductivity

To find the value of $x$, we step through the relationship between conductance, conductivity, molar conductivity, molarity, and mass percentage.

Step 1: Calculate the conductivity ($\kappa$) of the solution The conductivity ($\kappa$) is given by the product of conductance ($G$) and the cell constant $\left(\frac{l}{A}\right)$: $\kappa = G \times \left(\frac{l}{A}\right)$

Given:

• Conductance, $G = 1.9 \times 10^{-3} \text{ S}$
• Cell constant, $\frac{l}{A} = 1.3 \text{ cm}^{-1}$

$\kappa = (1.9 \times 10^{-3} \text{ S}) \times (1.3 \text{ cm}^{-1}) = 2.47 \times 10^{-3} \text{ S cm}^{-1}$

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Step 2: Calculate the molarity ($C$) of the solution The molar conductivity ($\Lambda_m$) is related to conductivity ($\kappa$) and molarity ($C$) by the equation: $\Lambda_m = \frac{\kappa \times 1000}{C}$

Given:

• Molar conductivity, $\Lambda_m = 123.5 \text{ S cm}^2 \text{ mol}^{-1}$

Substituting the known values: $123.5 = \frac{2.47 \times 10^{-3} \times 1000}{C}$

$123.5 = \frac{2.47}{C}$

$C = \frac{2.47}{123.5} = 0.02 \text{ mol L}^{-1}$

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Step 3: Relate molarity ($C$) to mass percentage ($x\% \text{ w/w}$) The formula relating molarity ($C$), mass percentage ($x\%$), density ($d$), and molar mass ($M$) of the solute is: $C = \frac{x \times d \times 10}{M}$

Given:

• Density of solution, $d = 1.0 \text{ g mL}^{-1}$
• Molar mass of $\text{MX}$, $M = 75 \text{ g mol}^{-1}$

Substitute the values into the formula: $0.02 = \frac{x \times 1.0 \times 10}{75}$

$0.02 = \frac{10x}{75}$

$10x = 0.02 \times 75$

$10x = 1.5$

$x = 0.15$

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Step 4: Express $x$ in the required format $x = 0.15 = 15 \times 10^{-2}$

The value of $x$ is $15 \times 10^{-2}$.