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NEET CHEMISTRYMedium

A 1 M solution of a compound 'X' has a density of 1.25 g/mL. If the molar mass of compound X is 85 g, what is the molality (m) of the solution?

A

0.705 m

B

1.208 m

C

1.165 m

D

0.858 m

Step-by-Step Solution

  1. Understand Definitions:
  • Molarity (M) = 1 mol/L [Class 12 Chemistry, Unit 1, Sec 1.2 (vi)]. This means there is 1 mole of solute 'X' in 1 Liter (1000 mL) of solution.
  • Density (d) = 1.25 g/mL.
  • Molar Mass (M_w) of X = 85 g/mol.

  1. Calculate Mass of Solution: Mass of solution=Volume×Density\text{Mass of solution} = \text{Volume} \times \text{Density} Mass=1000 mL×1.25 g/mL=1250 g\text{Mass} = 1000 \text{ mL} \times 1.25 \text{ g/mL} = 1250 \text{ g}

  2. Calculate Mass of Solute: Mass of solute=Moles×Molar Mass\text{Mass of solute} = \text{Moles} \times \text{Molar Mass} Mass=1 mol×85 g/mol=85 g\text{Mass} = 1 \text{ mol} \times 85 \text{ g/mol} = 85 \text{ g}

  3. Calculate Mass of Solvent: Mass of solvent=Mass of solution−Mass of solute\text{Mass of solvent} = \text{Mass of solution} - \text{Mass of solute} Mass=1250 g−85 g=1165 g=1.165 kg\text{Mass} = 1250 \text{ g} - 85 \text{ g} = 1165 \text{ g} = 1.165 \text{ kg}

  4. Calculate Molality (m): Molality (m)=Moles of soluteMass of solvent in kg\text{Molality (m)} = \frac{\text{Moles of solute}}{\text{Mass of solvent in kg}} [Class 12 Chemistry, Unit 1, Sec 1.2 (vii)] m=1 mol1.165 kg≈0.858 mol kg−1m = \frac{1 \text{ mol}}{1.165 \text{ kg}} \approx 0.858 \text{ mol kg}^{-1}

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