Sum
Calculate the amount of CaCl2 (molar mass = 111 g mol−1) which must be added to 500 g of water to lower its freezing point by 2 K, assuming CaCl2 is completely dissociated. (Kf for water = 1.86 K kg mol−1)
Advertisement Remove all ads
Solution
Given:
Molar mass of CaCl2 (MB) = 111 g/mol
Weight of water (wA) = 500 g
Kf for water = 1.86 K kg/mol
ΔTf = 2 K
Formula,
`DeltaT_f=ixx(K_fxxw_Bxx1000)/(w_AxxM_B)`
CaCl2 is an electrolyte which dissociates as,
CaCl2 → Ca2+ + 2Cl
Hence, i for CaCl2 = 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 CaCl2 required = 19.9 g
Concept: Colligative Properties and Determination of Molar Mass - Depression of Freezing Point
Is there an error in this question or solution?
APPEARS IN
Advertisement Remove all ads
