Question

If 4.85 g of Ethylene Glycol are added to 98.0 g of water what is the freezing point of the solution?

Answer 1

This question can be solved by using the colligative property, which is depression in freezing point . It is calculated as follows.

Given mass of ethylene glycol = 4.85 g

mass of water = 98.0 g

we know molecular weight of ethylene glycol = 62.07 g/mol

therefore molality of the given solution is calculated using the formula

therefore m = 4.85 x 1000/(98 x 62.07)

molality m = 0.797 m

We now can calculate the depression in freezing point by using the formula

T_f = K_f x m

where K_f is molal depression constant and for water K_f = 1.86 ⁰C/molal

therefore T_f = 1.86 x 0.797 ⁰C = 1.483 ⁰C

but T_f = freezing point of water - freezing point of solution

Therefore freezing point of solution = -1.48 ⁰C