Question 10 of 50intermediate🔧 ApplyNumerical2 marks

Boiling point of water at 750 mm Hg is 99.63°C. How much sucrose is to be added to 500 g of water such that it boils at 100°C.

Correct Answer

121.68 g (approx)

Exercise: Intext Questions | Q: 1.10 | (Page 23)
For More Understanding

Explanation

The problem requires calculating the mass of sucrose needed to elevate the boiling point of water. The elevation of boiling point (ΔTb) is the difference between the desired boiling point and the boiling point of pure water at the given pressure. The context explicitly provides the boiling point elevation formula and the value of Kb for water (0.52 K kg mol⁻¹). Although the molar mass of sucrose is not explicitly stated in the provided chunks, it is a standard value (342 g mol⁻¹) used in NCERT solutions for this specific textbook problem. The calculation involves rearranging the formula ΔTb = (Kb × 1000 × w₂) / (M₂ × w₁) to solve for the mass of sucrose (w₂).

Solution Steps

  1. Step 1: Calculate the elevation of boiling point (ΔTb\Delta T_b). The boiling point of pure water at 750 mm Hg is 99.63°C, and the desired boiling point is 100°C. Thus, ΔTb=Tb−Tb0=100−99.63=0.37 K\Delta T_b = T_b - T_b^0 = 100 - 99.63 = \mathbf{0.37\ K}.

  2. Step 2: Identify the known values from the context. The molal elevation constant (KbK_b) for water is given as 0.52 K kg mol⁻¹. The mass of solvent (w1w_1) is 500 g. The molar mass of sucrose (M2M_2) is 342 g mol⁻¹.

  3. Step 3: Use the formula for elevation of boiling point: ΔTb=Kb×1000×w2M2×w1\Delta T_b = \frac{K_b \times 1000 \times w_2}{M_2 \times w_1}. Rearrange to solve for mass of solute (w2w_2): w2=ΔTb×M2×w1Kb×1000w_2 = \frac{\Delta T_b \times M_2 \times w_1}{K_b \times 1000}.

  4. Step 4: Substitute the values into the equation: w2=0.37×342×5000.52×1000w_2 = \frac{0.37 \times 342 \times 500}{0.52 \times 1000}. Calculating this gives w2=63270520=121.68 gw_2 = \frac{63270}{520} = \mathbf{121.68\ g}.

PreviousNext