Question

\(250\) \(g\) solution of \(D-glucose\) in water contains 10.8% of carbon by weight. The molality of the solution is nearest to (Given: Atomic Weights are, \(H, 1 u; C, 12 u; O, 16 u\))

• A. 1.03
• B. 2.06
• C. 3.09
• D. 5.40

Answer 1

The Correct Option is B

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

1. Understand the composition of the solution: The solution contains 10.8% carbon by weight. The total weight of the solution is given as 250 g.
2. Calculate the amount of carbon in the solution: \( = \frac{10.8}{100} \times 250 = 27 \, \text{g of carbon}\)
3. Determine the amount of D-glucose in the solution:
   • The empirical formula of glucose is \(C_6H_{12}O_6\).
   • Molar mass of D-glucose = \( 6 \times 12 + 12 \times 1 + 6 \times 16 = 180 \, \text{g/mol} \).
   • Moles of carbon in glucose = \(\frac{27}{12} = 2.25 \, \text{moles}\).
   • Since each mole of glucose contains 6 moles of carbon, moles of glucose = \(\frac{2.25}{6} = 0.375 \, \text{moles}\).
   • Mass of D-glucose = 0.375 moles \(\times 180 \, \text{g/mol} = 67.5 \, \text{g}\).
4. Calculate the mass of water in the solution:

   Total mass = 250 g, mass of glucose = 67.5 g

   \(250 \, \text{g} - 67.5 \, \text{g} = 182.5 \, \text{g of water}\)

   Convert the mass of water to kilograms: \( 182.5 \text{ g} = 0.1825 \text{ kg}\)

5. Determine the molality of the solution:

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

   \(\text{Molality} = \frac{0.375 \, \text{moles of glucose}}{0.1825 \, \text{kg of water}} \approx 2.06 \, \text{mol/kg}\)

The molality of the solution is therefore nearest to 2.06, which is the correct answer.