The molar freezing point constant for water is 1.86° C / mol. If 342 g of cane sugar (C12 H22 O11) is dissolved in 1000 g of water, the solution will freeze at:

Question

The molar freezing point constant for water is 1.86° C / mol. If 342 g of cane sugar (C12 H22 O11) is dissolved in 1000 g of water, the solution will freeze at: (A)  -1.86°C  (B) 1.86°C  (C)  -3.92°C  (D) 2.42°C

Tardigrade Admin · Accepted Answer

Molality of cane sugar solution =(342/342 × 1)=1 m We know that, Δ Tf=Kf . m=1.86 × 1 m=1.86o Hence, freezing point of solution =0.00-(1.86)=-1.86° C

Q. The molar freezing point constant for water is $1.8 6^{\circ} C / m o l$ . If $342 g$ of cane sugar $(C_{12} H_{22} O_{11})$ is dissolved in $1000 g$ of water, the solution will freeze at:

$- 1.8 6^{\circ} C$

$1.8 6^{\circ} C$

$- 3.9 2^{\circ} C$

$2.4 2^{\circ} C$

Molality of cane sugar solution $= \frac{342}{342 \times 1} = 1 m$
We know that, $Δ T_{f} = K_{f} . m = 1.86 \times 1 m = 1.8 6^{o}$
Hence, freezing point of solution
$= 0.00 - (1.86) = - 1.8 6^{\circ} C$

Q. The molar freezing point constant for water is 1.86∘C/mol. If 342g of cane sugar (C12​H22​O11​) is dissolved in 1000g of water, the solution will freeze at:

−1.86∘C

1.86∘C

−3.92∘C

2.42∘C