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Radda [10]
3 years ago
6

A 2.4 m aqueous solution of an ionic compound with the formula mx2mx2 has a boiling point of 103.4 °c. calculate the van't hoff

factor (i) for mx2mx2 at this concentration.
Chemistry
1 answer:
saveliy_v [14]3 years ago
3 0

Answer:- Van't hoff factor(i) for the unknown ionic compound is 2.77.

Solution:- Elevation in boiling point is a colligative property and the equation used for solving this type of problems is...

delta Tb = i x m x kb

where delt Tb is elevation in boiling point.

i is Von't hoff factor.

m is molality of the solution and kb is the molal elevation constant. It's value for water is 0.512.

Boiling point of solution is given as 103.4 degree C and we know that water boils at 100 degree C.

So, delta Tb = 103.4 - 100 = 3.4 degree C

molality is given as 2.4m.

Let's plug in the values...

3.4 = i x 2.4 x 0.512

i = 3.4/(2.4 x 0.512)

i = 2.77

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