Molarity $(\mathrm{M})$ of an aqueous solution containing $x \mathrm{~g}$ of anhyd. $\mathrm{CuSO}_4$ in $500 \mathrm{~mL}$ solution at $32^{\circ} \mathrm{C}$ is $2 \times 10^{-1} \mathrm{M}$. Its molality will be _________ $\times 10^{-3} \mathrm{~m}$. (nearest integer). [Given density of the solution $=1.25 \mathrm{~g} / \mathrm{mL}$]

<p>To find the molality of the solution, we need to follow these steps:</p> <p>1. Determine the number of moles of anhydrous $\mathrm{CuSO}_4$ in the solution.</p> <p>The molarity (M) is given as $2 \times 10^{-1}$ M in a 500 mL solution. Hence, the number of moles of $\mathrm{CuSO}_4$ is:</p> <p>$$ \text{Moles of } \mathrm{CuSO}_4 = \text{Molarity} \times \text{Volume in liters} = 2 \times 10^{-1} \times 0.5 = 0.1 \text{ moles} $$</p> <p>2. Calculate the mass of the solution using the given density.</p> <p>The density of the solution is given as $1.25 \mathrm{~g/mL}$. The volume of the solution is 500 mL. Therefore, the mass of the solution is:</p> <p>$$ \text{Mass of solution} = \text{Density} \times \text{Volume} = 1.25 \times 500 = 625 \text{ grams} $$</p> <p>3. Find the mass of the solvent (water) in the solution by subtracting the mass of the solute (anhydrous $\mathrm{CuSO}_4$) from the total mass of the solution.</p> <p>We know the number of moles of $\mathrm{CuSO}_4$, and we can find its molar mass.</p> <p>$$ \text{Molar mass of } \mathrm{CuSO}_4 = 63.5 + 32 + 4 \times 16 = 159.5 \text{ g/mol} $$</p> <p>Therefore, the mass of $\mathrm{CuSO}_4$ is:</p> <p>$$ \text{Mass of } \mathrm{CuSO}_4 = \text{Number of moles} \times \text{Molar mass} = 0.1 \times 159.5 = 15.95 \text{ grams} $$</p> <p>4. Calculate the mass of the solvent (water):</p> <p>$$ \text{Mass of water} = \text{Mass of solution} - \text{Mass of } \mathrm{CuSO}_4 = 625 - 15.95 = 609.05 \text{ grams} $$</p> <p>Convert the mass of water to kilograms:</p> <p>$\text{Mass of water} = 609.05 \text{ grams} = 0.60905 \text{ kilograms}$</p> <p>5. Calculate the molality using the formula:</p> <p>$$ \text{Molality} (\text{m}) = \frac{\text{Moles of solute}}{\text{Mass of solvent in kg}} = \frac{0.1}{0.60905} = 0.164 \text{ mol/kg} $$</p> <p>Converting to the desired unit:</p> <p>$\text{Molality} (\text{m}) = 0.164 \times 10^3 = 164 \times 10^{-3}$</p> <p>Therefore, the molality of the solution is approximately 164 $\times 10^{-3} \mathrm{~m}$.</p>