<u>Answer:</u> The boiling point of solution is 100.53
<u>Explanation:</u>
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:
![\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of pure solution}](https://tex.z-dn.net/?f=%5CDelta%20T_b%3D%5Ctext%7BBoiling%20point%20of%20solution%7D-%5Ctext%7BBoiling%20point%20of%20pure%20solution%7D)
To calculate the elevation in boiling point, we use the equation:
![\Delta T_b=iK_bm](https://tex.z-dn.net/?f=%5CDelta%20T_b%3DiK_bm)
Or,
![\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)}}](https://tex.z-dn.net/?f=%5Ctext%7BBoiling%20point%20of%20solution%7D-%5Ctext%7BBoiling%20point%20of%20pure%20solution%7D%3Di%5Ctimes%20K_b%5Ctimes%20%5Cfrac%7Bm_%7Bsolute%7D%5Ctimes%201000%7D%7BM_%7Bsolute%7D%5Ctimes%20W_%7Bsolvent%7D%5Ctext%7B%20%28in%20grams%29%7D%7D)
where,
Boiling point of pure solution = 100°C
i = Vant hoff factor = 2 (For CsCl)
= molal boiling point elevation constant = 0.51°C/m
= Given mass of solute (CsCl) = 8.00 g
= Molar mass of solute (CsCl) = 168.4 g/mol
= Mass of solvent (water) = 92 g
Putting values in above equation, we get:
![\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](https://tex.z-dn.net/?f=%5Ctext%7BBoiling%20point%20of%20solution%7D-100%3D2%5Ctimes%200.51%5EoC%2Fm%5Ctimes%20%5Cfrac%7B8.00%5Ctimes%201000%7D%7B168.4g%2Fmol%5Ctimes%2092%7D%5C%5C%5C%5C%5Ctext%7BBoiling%20point%20of%20solution%7D%3D100.53%5EoC)
Hence, the boiling point of solution is 100.53