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If 1 gram of calcium chloride is dissolved in 100 mL of pure water, what temperature change will be observed

User Eli Burke
by
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1 Answer

4 votes

Step-by-step explanation:

It is known that the molecular weight of
CaCl_(2) is 111 g/mol. This means that 1 mole of
CaCl_(2) contains 111 g
CaCl_(2).

1 g
CaCl_(2) =
(1)/(111) mol CaCl_(2)

As we know that the density of water is 1 g/cc (as 1 ml = 1 cc).

So, 100 ml water = 100 g water. Therefore, in 100 g of water
CaCl_(2) present will be calculated as follows.


(1)/(111) mol

So, in 1000 g water the amount of
CaCl_(2) present will be calculated as follows.


(1 * 1000)/(111 * 100)

= 0.09 mol

Hence, the molality of
CaCl_(2) is 0.09 mol.

According to Raoult's law,


\Delta T_(b) = K_(b) * m

where,
K_(b) = boiling point constant

For pure 1 kg water,
K_(b) = 0.52 K.kg/mol

m = molality of solution

Therefore, putting the given values into the above formula as follows.


\Delta T_(b) = K_(b) * m

=
0.52 K m^(-1) * 0.09 m

= 0.0468 K

Therefore, the boiling point will raise by 0.0468 K.

User Tom Clarkson
by
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