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0.37°C

We know,

$∆$ t_f = i k_jm

where, $∆$ t_f = depression in freezing point; i = van't Hoff factor; m = molality and k_f = freezing point depression constant.

For 0.01 molal NaCl solution

0.37 = 2 × k_f × 0.01

$∴$ k_f = $\frac{0.37}{2 \times 0.01}$ ...(i)

For 0.02 molal urea solution

$∆$ t_f = 1 × k_f× 0.02

$∴$ t_f = $\frac{∆ t_{f}}{0.02}$ ...(ii)

$∆ t_{f} = \frac{0.37 \times 0.02}{2 \times 0.01}$

$∴$ $∆ t_{f} = 0.37 ° C$

602.

Chloroform, when kept open, is oxidised to :

603.

Blood cells will remain as such in:

604.

The mixture that forms maximum boiling azeotrope is:

605.

For an ideal solution, the correct option is :