Read the passage carefully and answer the Questions.

A homogeneous mixture of two or more than two components has uniform composition and properties. The binary solutions have two components, which may be solid, liquid or gas. The resultant solution has a larger amount of solvent and a smaller amount of solute. The amount of solute in a given amount of solvent decides the concentration of the resulting solution. The concentration can be expressed in various ways, like molarity, molality, normality and mole fraction.

The molality of a 1 liter aqueous solution of 93% $H_2SO_4$ (mass/volume) is:

(Given: density of the solution = $1.84\, g\, mL^{-1}$, molar mass of $H_2SO_4 = 98\, g\, mol^{-1}$)

$10.43\, mol\, kg^{-1}$

The correct answer is Option (1) → $10.43\, mol\, kg^{-1}$

• Solution volume: $1\text{ L} = 1000\text{ mL}$
• Concentration: $93\%$ (mass/volume)
• Density of solution ($\rho$): $1.84\text{ g/mL}$
• Molar mass of $H_2SO_4$: $98\text{ g/mol}$

 mass of the solute ($H_2SO_4$)

The $93\%$ (m/v) concentration means there are $93\text{ g}$ of solute in every $100\text{ mL}$ of solution.

For $1000\text{ mL}$ of solution:

$\text{Mass of solute } (w_2) = \frac{93}{100} \times 1000 = 930\text{ g}$

 Total mass of the solution

Using the density formula ($\text{Mass} = \text{Density} \times \text{Volume}$):

$\text{Total mass of solution} = 1.84\text{ g/mL} \times 1000\text{ mL} = 1840\text{ g}$

Based on the image provided, here is the content converted into LaTeX.

Mass of the solvent (water)

$\text{Mass of solvent } (w_1) = \text{Total mass} - \text{Mass of solute}$

$\text{Mass of solvent } (w_1) = 1840 \text{ g} - 930 \text{ g} = 910 \text{ g} = 0.91 \text{ kg}$

 Moles of Solute

$\text{Moles of } H_2SO_4(n) = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{930 \text{ g}}{98 \text{ g/mol}} \approx 9.4897 \text{ mol}$

 Molality ($m$)

Molality is defined as moles of solute per kilogram of solvent:

$m = \frac{\text{Moles of solute}}{\text{Mass of solvent in kg}}$

$m = \frac{9.4897 \text{ mol}}{0.91 \text{ kg}} \approx 10.428 \text{ mol/kg}$

The correct answer is Option 1: $10.43 \text{ mol kg}^{-1}$.