When 'x' × 10⁻² mL methanol (molar mass =32 g' density =0.792 g / cm³) is added to 100 mL. water (density =1 g / cm³), the following diagram is obtained.
x= ________ (nearest integer).
[Given : Molal freezing point depression constant of water at 273.15 K is 1.86 K kg mol⁻¹]

Solution

Chemistry ma’am · 4 steps

Read the freezing-point drop off the graph, then work back from molality to methanol's volume.

1. Read the graph. Pure water freezes at 273.15 K. The methanolic solution freezes where its curve meets the frozen-system curve, at 270.65 K. So Δ T_f = 2.5 K.

   Δ T_f = K_f m ⇒ m = 2.5/1.86 = 1.344 mol kg⁻¹

2. Moles and mass. 100 mL of water at 1 g cm⁻³ is 0.1 kg, so the moles of methanol are 1.344 × 0.1 = 0.1344 mol. Its mass is 0.1344 × 32 = 4.301 g.
3. Mass to volume. Volume = 4.301/0.792 = 5.431 mL = 543.1 × 10⁻² mL.
4. So. x ≈ 543.

The common slip: 430. Stopping at the mass of methanol and skipping the density step gives 4.30 g, that is, 430 instead of 543.

The idea to keep. On such a graph, the freezing point is where the solution curve meets the solid curve. From Δ T_f get molality, then moles, mass and finally volume through density.

Answer from the NTA final answer key. Solution by PrepWiser.