Question

1000 g of 1 m sucrose solution in water is cooled to −3.534°C. What weight of ice would be separated out at this temperature 1 is ______ gm. K_f(H₂O) = 1.86 K mol⁻¹ Kg)

Options

• 125

• 275

• 353

• 470

Answer 1

1000 g of 1 m sucrose solution in water is cooled to −3.534°C. What weight of ice would be separated out at this temperature 1 is 353 gm.

Explanation:

ΔT_f = `(1000 xx "K"_"f" xx w)/("W" xx "m")` ....(i)

= K_f × molality 

= 1.86 × 1

= 1.86 .....(ii)

Again from the above equation

`w/"W" = (Δ"T"_"f" xx "m")/("K"_"f" xx 1000)`

= `(1.86 xx 342)/(1.86 xx 1000)`

= 0.342 ....(iii)

Again `w` + W = 1000 ...(iv) [As the weight of solution = 1000 g]

From equations (iii) and (iv) we get

`w` = 254.84 g

W = 754.16 g

As the solution freezed up to –3.534, the weight of sucrose remains unchanged.

So, from (i) we get

3.534 = `(1.86 xx 1000 xx 254.84)/(342 xx "W"_1)`

W₁ = 392.18 g

∴ Ice separated = 745.16 – 392.18 = 352.98 gm = 353 gm