Question

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

Answer 1

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.