You visited us 1 times! Enjoying our articles? Unlock Full Access!

Depression in Freezing Point

QUESTION 2.20...

Question

QUESTION 2.20

A $5 %$ solution (by mass) of cane sugar in water has freezing point 271 K. Calculate freezing point of $5 %$ glucose in water if freezing point of pure water is 273.15K.

Open in App

Solution

For cane sugar solution
$W_{b} (s u g a r) = 5 g; W_{A} (w a t e r) = 100 - 5 = 95 g M_{B} (s u g a r) = 342 g m o l^{- 1}; Δ T_{f} = (273.15 - 271.00) K = 2.15 K Δ T_{f} = \frac{K_{f} \times W_{B}}{M_{B} \times W_{A}} (2.15 K) = \frac{K_{f} \times (5 g)}{(342 g m o l^{- 1}) \times (95 g)} \dots \dots (i)$
For glucose solution
$W_{B} (g l u c o s e) = 5 g; W_{A} (w a t e r) = 100 - 5 = 95 g M_{B} (g l u c o s e) = 180 g m o l^{- 1}; Δ T_{f} = ? Δ T_{f} = \frac{K_{f} \times (5 g)}{(180 g m o l^{- 1}) \times (95 g)} \dots \dots (i i)$
On dividing Eq. (ii) by Eq. (i)
$\frac{Δ T_{f}}{(2.15 K)} = \frac{K_{f} \times (5 g)}{(180 g m o l^{- 1}) \times (95 g)} \times \frac{(342 g m o l^{- 1}) \times (95 g)}{(K_{f}) \times (5 g)} Δ T_{f} = \frac{342 \times 2.15}{180} K = 4.085 K$
Freezing point temperature for 5\% glucose solution
= (273.15 - 4.085)K = 269.07K

Suggest Corrections

199

Similar questions

5

271 K

273.15 K

5

5

= 342 g

{m o l}^{- 1}

271 K

5

= 18 g

{m o l}^{- 1}

273.15 K

10

269.15 K

10

273.15 K

= 342 g m o l^{-}

= 180 g m o l^{- 1}

g m o l^{- 1})

g m o l^{- 1})

Join BYJU'S Learning Program

Explore more

Join BYJU'S Learning Program