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Standard XII

Chemistry

Depression in Freezing Point

Calculate the...

Question

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

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Solution

$C a C l_{2} \to C a^{2 +} + 2 C l^{-}$

$∴ i = 3$

$Δ T_{f} = i \times K_{f} \times m$

$2 = 3 \times 1.86 \times m$

$∴ m = \frac{2}{3 \times 1.86} = \frac{2}{5.58} = 0.358$

$∴$ Amount of $C a C l_{2}$ to be added to
500 g of water $= \frac{0.358 \times 500}{1000} \times 111$

=19.869 g.

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Similar questions

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

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 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})

Q. Calculate the depression in freezing point of water when

20.0 g

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

is added to

500 g

of water.
[Given:

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

Q. What mass of ethylene glycol (Molar mass

= 62.0

m o l^{- 1}

) must be added to

5.50

kg of water to lower the freezing point of water from

0^{o}

C to -

10^{o}

C ? (

K_{f}

for water

= 1.86

K Kg

m o l^{- 1}

)

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Calculate the amount of CaCl2 (molar mass =111gmol−1) which must be added to 500 g of water to lower its freezing point by 2K, assuming CaCl2 is completely dissociated. (Kf for water =1.86Kkg mol−1).

CaCl2→Ca2++2Cl− ∴i=3 ΔTf=i×Kf×m 2=3×1.86×m ∴m=23×1.86=25.58=0.358 ∴ Amount of CaCl2 to be added to 500 g of water =0.358×5001000×111 =19.869 g.

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

$C a C l_{2} \to C a^{2 +} + 2 C l^{-}$

$∴ i = 3$

$Δ T_{f} = i \times K_{f} \times m$

$2 = 3 \times 1.86 \times m$

$∴ m = \frac{2}{3 \times 1.86} = \frac{2}{5.58} = 0.358$

$∴$ Amount of $C a C l_{2}$ to be added to
500 g of water $= \frac{0.358 \times 500}{1000} \times 111$

=19.869 g.