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

Options

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

Answer

C

Explanation

Molarity (cxpcrimental)
` = (Delta T _(f))/(K _(f)) = (3.82)/( 1.86) =(3.82)/(1.86)= 2.054` mol/1000 g solvent
Molarity (thcartical) `= ("mole of solute")/("wt. of solventing(g))xx1000`
`= ( 5g//142 g //"mole")/(x) xx1000`
`Na_(2) SO_(4) to 2Na^(+) +SO_(4)^(2-)`
`{:("Moles before dissociation", 1,0,0),("Moles after dissociation", 1-x, 2x, x):}`
Von't Factor (i) `= ("Moles after dissociation")/("Moles before dissociation")`
`= ((1-x 0 + 2x+x)/( 1)`
`Na_(2) SO_(4)` is ionised `81.5%` means `x=0.815`
`=((1-0.815) + 2xx0.815 +0.815)/(1) =2.63.`
`i = ("Observed molarity ")/("Calculated molarity")`
`implies 2.63 =(2.054)/((0.0352)/(x)xx 1000) =45.07g.`