5 g of `Na_(2)SO_(4)` was dissolved in x g of `H_(2)O` . The change in freezing point was found to be `3.82^(@)C` . If `Na_(2)SO_(4)` is `81.5%` ionised , the value of x
(`k_(f)` for water =`1.86^(@)C` kg `"mol"^(-1)`) is apporximately :
(molar mass of S=32 g `"mol"^(-1)` and that of Na=23 g `"mol"^(-1)`)

Options

A. 15g
B. 25g
C. 45 g
D. 65 g

Answer

C

Explanation

Molarity (experimental)
`=(DeltaT_f)/(K_f)=(3.82)/(1.86) = 2.054 "mol/1000 g solvent"`
Molarity (theortical) `=("mole of solute")/("wt. of solvent(g)") xx 1000`
`=(5g//142g//"mole")/x xx1000`

Von.t Hoff Factor (i) `=("Moles after dissociation")/("Moles before dissociation")`
`=((1-x)+2x +x)/1`
`Na_2SO_4` is ionised 81.5%
means x = 0.815
`=((1-0.815) + 2xx 0.815 + 0.815)/1`
`=2.63`.
`i=("Observed molarity")/("Calculated molarity")`
`rArr 2.63 = (2.054)/((0.0253)/x xx 1000), x = 45.07 g`