The amount of urea to be dissolved in 500 cc of water (K=1.86° C ) to produce a depression of 0.186° C in the freezing point is :

Question

The amount of urea to be dissolved in 500 cc of water (K=1.86° C ) to produce a depression of 0.186° C in the freezing point is : (A) 9 g (B) 6 g (C) 3 g (D) 0.3 g

Tardigrade Admin · Accepted Answer

We know that, w (mass of solute) =(m × Δ Tf × W/1000 × Kf) m=mol. wt. of urea (60) Δ Tf=0.186° C Kf=1.86°, W=500 g =(60 × 0.186 × 500/1000 × 1.86) =3 g

Q. The amount of urea to be dissolved in $500 cc$ of water $(K = 1.8 6^{\circ} C)$ to produce a depression of $0.18 6^{\circ} C$ in the freezing point is :

9 g

6 g

3 g

0.3 g

We know that,
w (mass of solute) $= \frac{m \times Δ T _{f} \times W}{1000 \times K _{f}}$
m=mol. wt. of urea (60)
$Δ T_{f} = 0.18 6^{\circ} C$
$K_{f} = 1.8 6^{\circ}, W = 500 g$
$= \frac{60 \times 0.186 \times 500}{1000 \times 1.86}$
$= 3 g$

Q. The amount of urea to be dissolved in 500cc of water (K=1.86∘C) to produce a depression of 0.186∘C in the freezing point is :

9 g

6 g

3 g

0.3 g

We know that, w (mass of solute) =1000×Kf​m×ΔTf​×W​ m=mol. wt. of urea (60) ΔTf​=0.186∘C Kf​=1.86∘,W=500g =1000×1.8660×0.186×500​ =3g

Q. The amount of urea to be dissolved in $500 cc$ of water $(K = 1.8 6^{\circ} C)$ to produce a depression of $0.18 6^{\circ} C$ in the freezing point is :

We know that,
w (mass of solute) $= \frac{m \times Δ T _{f} \times W}{1000 \times K _{f}}$
m=mol. wt. of urea (60)
$Δ T_{f} = 0.18 6^{\circ} C$
$K_{f} = 1.8 6^{\circ}, W = 500 g$
$= \frac{60 \times 0.186 \times 500}{1000 \times 1.86}$
$= 3 g$