JEE Main 27-Aug-2021 Shift 1 Solved Paper

Let mass of water (initially present) = x g
Mass of sucrose =(1000−x)g
Moles of sucrose =

1000−x

342

Molality =

moles‌of‌sucrose

mass‌of‌water(initially)

0.75=

( 1000−x 342 )

( x 1000 )

x

1000

=

1000−x

342×0.75

256.5x=106−1000x
⇒ x = 795.86 g
Moles of sucrose = 0.5969
New mass of H2O=a‌kg
Depression in freezing point ΔTf=Kf×m
4=

0.5969

a

×1.86
⇒ a = 0.2775 kg
Ice separated =795.86−277.5
= 518.3 g ≈ 518 g.