Question

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

Answer 1

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 $CaCl_{2}$ = $\frac{1}{111} mol CaCl_{2}$

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.

              $\frac{1}{111}$ mol

So, in 1000 g water the amount of $CaCl_{2}$ present will be calculated as follows.

                   $\frac{1 \times 1000}{111 \times 100}$

                  = 0.09 mol

Hence, the molality of $CaCl_{2}$ is 0.09 mol.

According to Raoult's law,

            $\Delta T_{b} = K_{b} \times m$

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.

               $\Delta T_{b} = K_{b} \times m$

                          = $0.52 K m^{-1} \times 0.09 m$

                          = 0.0468 K

Therefore, the boiling point will raise by 0.0468 K.