A solution of sucrose (molar mass =342 gmol- 1 ) has been prepared by dissolving 68.5 g of sucrose in 1000 g of water. The freezing point of the solution obtained will be ( Kf for water =1.86 Kkgmol- 1 )

Question

A solution of sucrose (molar mass =342 gmol- 1 ) has been prepared by dissolving 68.5 g of sucrose in 1000 g of water. The freezing point of the solution obtained will be ( Kf for water =1.86 Kkgmol- 1 ) (A) -0.570 C (B) -0.372 C (C) -0.520 C (D) +0.372 C

Tardigrade Admin · Accepted Answer

Δ Tf=Kf.(w/m o l . w t . × Ws o l .)× 1000 Δ Tf=((68.5 × 1000/342 × 1000))× 1.86 Δ Tf=0.372 F.pt. of solution =0oC-0.372=-0.372C

Q. A solution of sucrose (molar mass $= 342 g m o l^{- 1}$ ) has been prepared by dissolving 68.5 g of sucrose in 1000 g of water. The freezing point of the solution obtained will be

( $K_{f}$ for water $= 1.86 K k g m o l^{- 1}$ )

-0.570 C

-0.372 C

-0.520 C

+0.372 C

$Δ T_{f} = K_{f} . \frac{w}{m o l . wt . \times W _{so l .}} \times 1000$

$Δ T_{f} = (\frac{68.5 \times 1000}{342 \times 1000}) \times 1.86$

$Δ T_{f} = 0.372$

F.pt. of solution $= 0^{o} C - 0.372 = - 0.372 C$

Q. A solution of sucrose (molar mass =342gmol−1 ) has been prepared by dissolving 68.5 g of sucrose in 1000 g of water. The freezing point of the solution obtained will be ( Kf​ for water =1.86Kkgmol−1 )