Calculate the temperature at which the solution of $54 g$ glucose present in $250 g$ water will be freezed. $(K_{f} = 1.86 K k g {m o l}^{- 1})$

Open in App

Solution

Given that:
Weight of solute $(W_{B}) = 54 g$

Molecular weight of solute $(M_{B}) = 180 g$ molWeight of solvent $(W_{A}) = 250 g$

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

Depression in freezing point $Δ T_{f}$

$= \frac{K_{f} \times W_{B} \times 1000}{M_{B} \times W_{A}}$

$= \frac{1.86 \times 54 \times 1000}{180 \times 250} = 2.23 K$

Depression in freezing point $= 2.23 K$

Freezing point of solution $= ?$

Freezing point of water $= 273.15 K$

$(Δ T_{f}) =$ Freezing point of water - Freezing point of solution

Freezing point of solution $= 273.15 - 2.23 = 270.92 K$

Suggest Corrections

Similar questions

Q. Calculate the depression in the freezing point of water when

10 g

C H_{3} C H_{2} C H C l C O O H

is added to

250 g

of water.

[

K a = 1.4 \times 10^{- 3}, K_{f} = 1.86 K k g m o l^{- 1}

]

Question 2.32

Calculate the depression in the freezing point of water when 10g of $C H_{3} C H_{2} C H C l C O O H$ is added to 250 g of water, $K_{a} = 1.4 \times 10^{- 3}$ ; $K_{f} = 1.86 K k g m o l^{- 1}$

0.1 m

urea solutions will have freezing point

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

Q. Calculate the freezing point of a solution containing

60 g

of glucose (Molar mass

= 180 g m o l^{- 1}

) in

250 g

of water.

(

K f

of water

= 1.86 K k g m o l^{- 1}

Calculate the temperature at which a solution containing 54 g of glucose, C₆H₁₂O₆ in 250 g of water will freeze. (Kf for water = 1.86 K kg mol^-1)

Join BYJU'S Learning Program

Calculate the temperature at which the solution of 54g glucose present in 250g water will be freezed. (Kf=1.86Kkgmol−1)

Calculate the temperature at which the solution of $54 g$ glucose present in $250 g$ water will be freezed. $(K_{f} = 1.86 K k g {m o l}^{- 1})$