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QJEEM24Ch2004: JEE Main 2024 Chemistry - Class 11 JEEM24Ch2004 Solution: Molarity of an aqueous solution containing of anhyd. in…

Molarity (M)(\mathrm{M}) of an aqueous solution containing x gx \mathrm{~g} of anhyd. CuSO4\mathrm{CuSO}_4 in 500 mL500 \mathrm{~mL} solution at 32∘C32^{\circ} \mathrm{C} is 2×10−1M2 \times 10^{-1} \mathrm{M} . Its molality will be _________ ×10−3 m\times 10^{-3} \mathrm{~m} . (nearest integer). [Given density of the solution =1.25 g/mL=1.25 \mathrm{~g} / \mathrm{mL} ]

Step-by-Step Verified Solution

To find the molality of the solution, we need to follow these steps:

1. Determine the number of moles of anhydrous CuSO4\mathrm{CuSO}_4 in the solution.

The molarity (M) is given as 2×10−12 \times 10^{-1} M in a 500 mL solution. Hence, the number of moles of CuSO4\mathrm{CuSO}_4 is:

Moles of CuSO4=Molarity×Volume in liters=2×10−1×0.5=0.1 moles\text{Moles of } \mathrm{CuSO}_4 = \text{Molarity} \times \text{Volume in liters} = 2 \times 10^{-1} \times 0.5 = 0.1 \text{ moles}

2. Calculate the mass of the solution using the given density.

The density of the solution is given as 1.25 g/mL1.25 \mathrm{~g/mL} . The volume of the solution is 500 mL. Therefore, the mass of the solution is:

Mass of solution=Density×Volume=1.25×500=625 grams\text{Mass of solution} = \text{Density} \times \text{Volume} = 1.25 \times 500 = 625 \text{ grams}

3. Find the mass of the solvent (water) in the solution by subtracting the mass of the solute (anhydrous CuSO4\mathrm{CuSO}_4 ) from the total mass of the solution.

We know the number of moles of CuSO4\mathrm{CuSO}_4 , and we can find its molar mass.

Molar mass of CuSO4=63.5+32+4×16=159.5 g/mol\text{Molar mass of } \mathrm{CuSO}_4 = 63.5 + 32 + 4 \times 16 = 159.5 \text{ g/mol}

Therefore, the mass of CuSO4\mathrm{CuSO}_4 is:

Mass of CuSO4=Number of moles×Molar mass=0.1×159.5=15.95 grams\text{Mass of } \mathrm{CuSO}_4 = \text{Number of moles} \times \text{Molar mass} = 0.1 \times 159.5 = 15.95 \text{ grams}

4. Calculate the mass of the solvent (water):

Mass of water=Mass of solution−Mass of CuSO4=625−15.95=609.05 grams\text{Mass of water} = \text{Mass of solution} - \text{Mass of } \mathrm{CuSO}_4 = 625 - 15.95 = 609.05 \text{ grams}

Convert the mass of water to kilograms:

Mass of water=609.05 grams=0.60905 kilograms\text{Mass of water} = 609.05 \text{ grams} = 0.60905 \text{ kilograms}

5. Calculate the molality using the formula:

Molality(m)=Moles of soluteMass of solvent in kg=0.10.60905=0.164 mol/kg\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}

Converting to the desired unit:

Molality(m)=0.164×103=164×10−3\text{Molality} (\text{m}) = 0.164 \times 10^3 = 164 \times 10^{-3}

Therefore, the molality of the solution is approximately 164 ×10−3 m\times 10^{-3} \mathrm{~m} .

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