Sum

Calculate the amount of CaCl_{2} (molar mass = 111 g mol^{−1}) which must be added to 500 g of water to lower its freezing point by 2 K, assuming CaCl_{2} is completely dissociated. (K_{f} for water = 1.86 K kg mol^{−1})

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#### Solution

Given:

Molar mass of CaCl_{2} (M_{B}) = 111 g/mol

Weight of water (wA) = 500 g

Kf for water = 1.86 K kg/mol

ΔT_{f} = 2 K

Formula,

`DeltaT_f=ixx(K_fxxw_Bxx1000)/(w_AxxM_B)`

CaCl_{2} is an electrolyte which dissociates as,

CaCl_{2 }→ Ca^{2+} + 2Cl

Hence, i for CaCl_{2} = 3

Solution:

`DeltaT_f=ixx(K_fxxw_Bxx1000)/(w_AxxM_B)`

`2 =3xx(1.86xxw_Bxx1000)/(500xx111)`

`w_B=(2xx500xx111)/(3xx1.86xx1000)`

`w_B=19.9 g`

Amount of CaCl_{2} required = 19.9 g

Concept: Colligative Properties and Determination of Molar Mass - Depression of Freezing Point

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