Respuesta :
Explanation:
It is known that the molecular weight of [tex]CaCl_{2}[/tex] is 111 g/mol. This means that 1 mole of [tex]CaCl_{2}[/tex] contains 111 g [tex]CaCl_{2}[/tex].
1 g [tex]CaCl_{2}[/tex] = [tex]\frac{1}{111} mol CaCl_{2}[/tex]
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 [tex]CaCl_{2}[/tex] present will be calculated as follows.
[tex]\frac{1}{111}[/tex] mol
So, in 1000 g water the amount of [tex]CaCl_{2}[/tex] present will be calculated as follows.
[tex]\frac{1 \times 1000}{111 \times 100}[/tex]
= 0.09 mol
Hence, the molality of [tex]CaCl_{2}[/tex] is 0.09 mol.
According to Raoult's law,
[tex]\Delta T_{b} = K_{b} \times m[/tex]
where, [tex]K_{b}[/tex] = boiling point constant
For pure 1 kg water, [tex]K_{b}[/tex] = 0.52 K.kg/mol
m = molality of solution
Therefore, putting the given values into the above formula as follows.
[tex]\Delta T_{b} = K_{b} \times m[/tex]
= [tex]0.52 K m^{-1} \times 0.09 m[/tex]
= 0.0468 K
Therefore, the boiling point will raise by 0.0468 K.
