Answer :
Answer: The boiling point of solution is 100.53
Explanation:
We are given:
8.00 wt % of CsCl
This means that 8.00 grams of CsCl is present in 100 grams of solution
Mass of solvent = (100 - 8) g = 92 grams
The equation used to calculate elevation in boiling point follows:
[tex]\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of pure solution}[/tex]
To calculate the elevation in boiling point, we use the equation:
[tex]\Delta T_b=iK_bm[/tex]
Or,
[tex]\text{Boiling point of solution}-\text{Boiling point of pure solution}=i\times K_b\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}[/tex]
where,
Boiling point of pure solution = 100°C
i = Vant hoff factor = 2 (For CsCl)
[tex]K_b[/tex] = molal boiling point elevation constant = 0.51°C/m
[tex]m_{solute}[/tex] = Given mass of solute (CsCl) = 8.00 g
[tex]M_{solute}[/tex] = Molar mass of solute (CsCl) = 168.4 g/mol
[tex]W_{solvent}[/tex] = Mass of solvent (water) = 92 g
Putting values in above equation, we get:
[tex]\text{Boiling point of solution}-100=2\times 0.51^oC/m\times \frac{8.00\times 1000}{168.4g/mol\times 92}\\\\\text{Boiling point of solution}=100.53^oC[/tex]
Hence, the boiling point of solution is 100.53