Question

If $18 \,g$ of glucose is dissolved is $250\, g$ of water then freezing point of the solution will be ($K_f$ of water $=1.86\, K \,kg\, mol ^{-1}$ )

• A. $-1.74^{\circ} C$
• B. $-0.25^{\circ} C$
• C. $-0.74^{\circ} C$
• D. $-1.3^{\circ} C$

Answer 1

Answer: (C)

$\Delta T _{ f }= mK _{ f }$
$=\frac{18 / 180}{250} \times 1000 \times 1.86$
$\Delta T _{ f }=0.74$
$T _{ f }=-0.74^{\circ} C$