Calculate the amount of $K C l$ which must be added to 1 kg of water so that the freezing point is depressed by 2K. $(k_{f}$ for water $= 1.86 K k g m o l^{- 1})$

Open in App

Solution

Change in freezing point, $Δ T_{f} = k_{f} \times m$
m= $\frac{Δ T_{f}}{k_{f}}$
$= \frac{2}{1.86} = 1.075 m o l e / k g$
$m = \frac{n u m b e r o f m o l e s o f s o l u t e}{m a s s o f s o l v e n t}$
$⟹ 1.075 = \frac{n}{1}$
$⟹ n = 1.075$

Suggest Corrections

Similar questions

Q. Calculate the amount of

C a C l_{2}

(molar mass

= 111 g m o l^{- 1})

which must be added to 500 g of water to lower its freezing point by 2K, assuming

C a C l_{2}

is completely dissociated.

(K_{f}

for water

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

Calculate the amount of $C a C l_{2}$ (molar mass = 111 g $m o l^{- 1}$ ) which must be added to 500 g of water to lower its freezing point by 2 K, assuming $C a C l_{2}$ is completely dissociated. ( $K_{f}$ for water = 1.86 K kg $m o l^{- 1}$ )

Q. Calculate the amount of sodium chloride which must be added to one kilogram of water so that the freezing point of water is depressed by 3 K. (Given,

K_{f}

=1.86 K kg

m o l^{- 1}

, atomic mass of Na=23, Cl=35.5).

Q. Calculate the amount of

C a C l_{2}

(molar mass = 111g/mol) which must be added to 580g of water so that freezing point lower by 2K, assuming

C a C l_{2}

is completely dissociated:

[

K_{f}

for water = 1.86 K kg/mol].

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}

]

Join BYJU'S Learning Program

Calculate the amount of KCl which must be added to 1 kg of water so that the freezing point is depressed by 2K. (kf for water =1.86Kkgmol−1)

Change in freezing point, ΔTf=kf×mm=ΔTfkf=21.86=1.075 mole/kgm=number of moles of solutemass of solvent⟹1.075=n1⟹n=1.075

Calculate the amount of $K C l$ which must be added to 1 kg of water so that the freezing point is depressed by 2K. $(k_{f}$ for water $= 1.86 K k g m o l^{- 1})$

Change in freezing point, $Δ T_{f} = k_{f} \times m$
m= $\frac{Δ T_{f}}{k_{f}}$
$= \frac{2}{1.86} = 1.075 m o l e / k g$
$m = \frac{n u m b e r o f m o l e s o f s o l u t e}{m a s s o f s o l v e n t}$
$⟹ 1.075 = \frac{n}{1}$
$⟹ n = 1.075$