1.00 g of a non-electrolyte solute dissolved in 50.0 g of benzene lowers its freezing point by 0.40 K. What is the molar mass of the solute? (K_f for benzene = 5.12 K kg mol^-1)
This is half the correct value, as if ΔT_f were 0.80 K.
C.
This is double the correct value, as if the solvent mass were 25 g.
D.
This is a quarter of the correct value and does not follow from the formula.
NCERT: Class 12 Chemistry, Chapter 1, 1.6
Sample 2NEET level
1.00 g of a non-electrolyte solute dissolved in 50 g of benzene lowered its freezing point by 0.40 K. The freezing point depression constant of benzene is 5.12 K kg mol⁻¹. The molar mass of the solute is
A
256 g mol⁻¹
B
128 g mol⁻¹
C
512 g mol⁻¹
D
102.4 g mol⁻¹
Show the answer
The answer is A.
M2 = (Kf × w2 × 1000) / (ΔTf × w1). M2 = (5.12 K kg mol⁻¹ × 1.00 g × 1000 g kg⁻¹) / (0.40 K × 50 g) = 5120 / 20 = 256 g mol⁻¹.
B.
Half the correct value, as if ΔTf were taken as 0.80 K.
C.
Double the correct value, as if ΔTf were taken as 0.20 K.
D.
Leaves out ΔTf: 5.12 × 1000 / 50 = 102.4.
NCERT: Class 12 Chemistry, Unit 1 (Solutions), Colligative Properties and Determination of Molar Mass: Depression of Freezing Point
Practise Molecular mass from colligative properties
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