The molar freezing point constant for water is 1.86 o textC/mol text. If 342 g of cane sugar (C12H22O11) is dissolved in 1000 g of water, the solution will freeze at:

Question

The molar freezing point constant for water is 1.86 o textC/mol text. If 342 g of cane sugar (C12H22O11) is dissolved in 1000 g of water, the solution will freeze at: (A)  -1.86 oC  (B)  1.86 oC  (C)  -3.92 oC  (D)  2.42 oC

Tardigrade Admin · Accepted Answer

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

Q. The molar freezing point constant for water is $1.86^{o} C/mol .$ If 342 g of cane sugar $(C_{12} H_{22} O_{11})$ is dissolved in 1000 g of water, the solution will freeze at:

$- 1.86^{o} C$

$1.86^{o} C$

$- 3.92^{o} C$

$2.42^{o} C$

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

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

−1.86oC

1.86oC

−3.92oC

2.42oC