Glucose is added to 1 litre water to such an extent that (Δ T f / H f ) becomes equal to (1/1000), the weight of glucose added is:

Question

Glucose is added to 1 litre water to such an extent that (Δ T f / H f ) becomes equal to (1/1000), the weight of glucose added is: (A) 0.32 g (B) 0.42 g (C) 0.22 g (D) 0.18 g

Tardigrade Admin · Accepted Answer

Δ Tf=Kf × molality =Kf × (w × 1000/m × W) ∴ (Δ Tf/Kf)=(w × 1000/m × W) or (1/1000)=(w × 1000/180 × 1000) ∴ w=0.18 g

Q. Glucose is added to 1 litre water to such an extent that $\frac{Δ T _{f}}{H _{f}}$ becomes equal to $\frac{1}{1000}$ , the weight of glucose added is:

$0.32 g$

$0.42 g$

$0.22 g$

$0.18 g$

$Δ T_{f} = K_{f} \times$ molality $= K_{f} \times \frac{w \times 1000}{m \times W}$
$∴ \frac{Δ T _{f}}{K _{f}} = \frac{w \times 1000}{m \times W}$
or $\frac{1}{1000} = \frac{w \times 1000}{180 \times 1000} ∴ w = 0.18 g$

Q. Glucose is added to 1 litre water to such an extent that Hf​ΔTf​​ becomes equal to 10001​, the weight of glucose added is:

0.32g

0.42g

0.22g

0.18g