Coolent used in car radiator is aqueous solution of ethylene glycol. In order to prevent the solution from…
- 50
- 52
- 62
- 60
Solution
$\Delta \mathrm{T}_{\mathrm{f}}=0.3^{\circ} \mathrm{C}=\frac{\mathrm{K}_{\mathrm{f}} \times \mathrm{W}_{\mathrm{B}} \times 1000}{\mathrm{M}_{\mathrm{B}} \times \mathrm{W}_{\mathrm{A}}}$
$0.3=\frac{1.86 \times \mathrm{W}_{\mathrm{B}} \times 1000}{62 \times 5000}$
$\therefore \quad \mathrm{W}_{\mathrm{B}}=50 \mathrm{~g}$
The amount used should be more than $50 \mathrm{~g}$
Asked in: JEE-TOPICTESTS-CHEMISTRY