1.80 g of solute A was dissolved in 62.5 cm 3 of ethanol and freezing point of the solution was found to be 155.1 K. The molar mass of solute A is g mol -1. [Given: Freezing point of ethanol is 156.0 K. Density of ethanol is 0.80 g cm -3. Freezing point depression constant of ethanol is .2.00 K kg mol -1]

Question

1.80 g of solute A was dissolved in 62.5 cm 3 of ethanol and freezing point of the solution was found to be 155.1 K. The molar mass of solute A is g mol -1. [Given: Freezing point of ethanol is 156.0 K. Density of ethanol is 0.80 g cm -3. Freezing point depression constant of ethanol is .2.00 K kg mol -1] (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Mass of C 2 H 5 OH =62.5 × 0.8=50 g Δ T f = K f × m 0.9=2 × (1.8 × 1000/ M w × 50) M w =(2 × 1.8 × 1000/0.9 × 50)=80

Mass of C2​H5​OH=62.5×0.8=50g ΔTf​=Kf​×m 0.9=2×Mw​×501.8×1000​ Mw​=0.9×502×1.8×1000​=80

Mass of $C_{2} H_{5} O H = 62.5 \times 0.8 = 50 g$
$Δ T_{f} = K_{f} \times m$
$0.9 = 2 \times \frac{1.8 \times 1000}{M _{w} \times 50}$
$M_{w} = \frac{2 \times 1.8 \times 1000}{0.9 \times 50} = 80$