If 15 g of a solute in 100 g of water makes the solution freeze at -1° C , then 30 g. of the same solute in 100 g of water will make the solution freeze at:

Question

If 15 g of a solute in 100 g of water makes the solution freeze at -1° C , then 30 g. of the same solute in 100 g of water will make the solution freeze at: (A)  - 2° C  (B)  - 0.5° C  (C)  0° C  (D)  2° C

Tardigrade Admin · Accepted Answer

We know that, m=(1000Kf.w/Δ T⋅ W) (As the solute and solvent are same in both cases), hence: (W1/W1)=Δ T1 and (W2/W2)=Δ T2 or (W1⋅ W2/W1⋅ W2)=(Δ T1/Δ T2) (15× 100/100× 30)=([(0)-(-1)]/Δ T2) Δ T2=2× 1=2° Hence, now water will freeze at (0-2)=-2° C

Q. If 15 g of a solute in 100 g of water makes the solution freeze at $- 1^{\circ} C$ , then 30 g. of the same solute in 100 g of water will make the solution freeze at:

$- 2^{\circ} C$

$- 0.5^{\circ} C$

$0^{\circ} C$

$2^{\circ} C$

Q. If 15 g of a solute in 100 g of water makes the solution freeze at −1∘C , then 30 g. of the same solute in 100 g of water will make the solution freeze at:

−2∘C

−0.5∘C

0∘C

2∘C

Q. If 15 g of a solute in 100 g of water makes the solution freeze at $- 1^{\circ} C$ , then 30 g. of the same solute in 100 g of water will make the solution freeze at:

$- 2^{\circ} C$

$- 0.5^{\circ} C$

$0^{\circ} C$

$2^{\circ} C$