asked 161k views
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Plz help me with this problem what is the boiling point (in celsius) of a 1.56m aqueous solution of CaCl2?

2 Answers

6 votes

The answer is 102.39* C


Hope this helps



answered
User Fredrik LS
by
8.7k points
3 votes

Answer:

102.4 °C

Step-by-step explanation:

There are four colligative properties :

a) elevation in boiling point

b) depression in freezing point

c) relative lowering of vapor pressure

d) osmotic pressure

When we add a non volatile solute in a solvent, the solution has higher boiling point as compared to the pure solvent, this is known as elevation in boiling point.

The elevation in boiling point is related to molality as:

elevation in boiling point =i X KbXmolality

Kb (water) = 0.512

molality given as = 1.56 m

i = vant hoff factor =3 (for calcium chloride)

Putting values

elevation in boiling point = 3X0.512X1.56 = 2.4

Thus the boiling point of aqueous solution will be = 100+2.4 = 102.4 °C

answered
User Hywak
by
8.1k points

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