MyQuestionIcon

2

You visited us 2 times! Enjoying our articles? Unlock Full Access!

Depression in Freezing Point

1.00 g of a n...

Question

1.00 g of a non-electrolyte solute $(m o l a r m a s s 250 g m o l^{- 1})$ was dissolved in 51.2 g of benzene. If the freezing point depression constant, $k_{f}$ of benzene is 5.12 K kg mol^{-1}\), the freezing point of benzene will be lowered by

Open in App

Solution

From the relation, we have
$Δ T_{f} = k_{f} \times W_{2} \times \frac{1000}{W_{1}} \times M_{2}$

Suggest Corrections

thumbs-up

0

Similar questions

Q. 1 g of non-electrolyte solute dissolved in 50g of benzene lowered the freezing point of benzene by 0.40K. The freezing point depression constant of benzene is 5.12 K kg/mol. Find the molar mass of the solute.

Q. (a) State Raoult's law for a solution containing volatile components. How does Raoult's law become a special cause of Henry's law ? .
(b) 1.00 g of a non-electrolyte solute dissolved in 50 g of benzene lowered the freezing point of benzene by 0.40 K. Find the molar mass of the solute. (

K_{f}

5.12 K k g m o l^{- 1})

Q. The freezing point depression constant

(K_{f})

{5.12 K kg mol}^{- 1}

. The freezing point depression for the solution of molality

0.01 m

containing a non-electrolyte solute in benzene is
(rounded off upto two decimal places)

1.00

g of non-electrolyte solute dissolved in

50

g of benzene lowered the point of benzene by

0.40

K. The freezing point depression constant of benzene is

5.12

K kg/mol. Find the molar mass of the solute.

Q. Calculate the mass of compound (molar mass =

256 g {m o l}^{- 1}

) to be dissolved in

75 g

of benzene to lower its freezing point by

0.48 K

K_{f} = 5.12 K k g {m o l}^{- 1}

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Depression in Freezing Point

CHEMISTRY

Watch in App

explore_more_icon

Explore more

Depression in Freezing Point

Standard XII Chemistry

Join BYJU'S Learning Program

CrossIcon