The molar freezing point constant for water is 1.86° C/mole . 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° C/mole . If 342 g of cane sugar (C12H22O11) 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 =1.86° Hence, freezing point of solution =0.00-(1.86)=-1.86° C

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

$- 1.86^{\circ} C$

$1.86^{\circ} C$

$- 3.92^{\circ} C$

$2.42^{\circ} C$

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

Q. The molar freezing point constant for water is 1.86∘C/mole . 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