The freezing point of water is depressed by 0.37° C in a 0.01 molal NaCl solution. The freezing point of 0.02 molal solution of urea is depressed by

Question

The freezing point of water is depressed by 0.37° C in a 0.01 molal NaCl solution. The freezing point of 0.02 molal solution of urea is depressed by (A) 0.37° C (B) 0.74° C (C) 0.185° C (D) 0° C

Tardigrade Admin · Accepted Answer

As we know that, Δ tf=i kj m where, Δ tf= depression in freezing point i = van't - Hoff factor m= molality and kf= freezing point depression constant. For 0.01 molal NaCl solution 0.37 =2 × kf × 0.01 ∴ kf =(0.37/2 × 0.01) ...(i) For 0.02 molal urea solution Δ tf=1 × kf × 0.02 ∴ kf=(Δ tf/0.02) ...(ii) From Eqs (i) and (ii) (0.37/2 × 0.01) =(Δ tf/0.02) Δ tf =(0.37 × 0.02/2 × 0.01) ∴ Δ tf =0.37° C

Q. The freezing point of water is depressed by $0.3 7^{\circ} C$ in a $0.01$ molal NaCl solution. The freezing point of $0.02$ molal solution of urea is depressed by

$0.3 7^{\circ} C$

$0.7 4^{\circ} C$

$0.18 5^{\circ} C$

$0^{\circ} C$

Q. The freezing point of water is depressed by 0.37∘C in a 0.01 molal NaCl solution. The freezing point of 0.02 molal solution of urea is depressed by

0.37∘C

0.74∘C

0.185∘C

0∘C

Q. The freezing point of water is depressed by $0.3 7^{\circ} C$ in a $0.01$ molal NaCl solution. The freezing point of $0.02$ molal solution of urea is depressed by

$0.3 7^{\circ} C$

$0.7 4^{\circ} C$

$0.18 5^{\circ} C$

$0^{\circ} C$