Question

An antifreeze solution is prepared by dissolving 31 g of ethylene glycol
(Molar mass = 62 g mol-?) in 600 g of water. Calculate the freezing point
of the solution. (K for water = 1.86 K kg mol-?)

Answer 1

Answer:

-1.55°C

Explanation:

well, the question asked in CBSE CHEMISTRY 2020.

here, molality = no. of moles of solute/ weight of kg in solvent .

Hence, m =

$\frac{ \frac{31}{62} }{0.6}$

= 5/6.

now, depression in freezing point = Kf × m

=1.86 × 5/6

= 1.55.

thus, freezing point of solution = 0°C - 1.55°C

= -1.55°C Ans

Regards

Kshitij Ampearl

Answer 2

Given:

w = 31 g

M.wt = 62 g/mol

W = 600 g

Kf = 1.86 K kg/mol

To Find:

Freezing point of the solution.

Calculation:

- The depression in freezing point of a solution can be given by the formula:

ΔTf = Kf × m

⇒ (Tf - T'f) = Kf × (w × 1000)/(M.wt × W)

⇒ (0 - T'f) = 1.86 × (31 × 1000)/(62 × 600)

⇒ -T'f = 1.55

⇒ T'f = -1.55 °C

- So the freezing point  of the solution is -1.55 °C.