An aqueous CsCl solution is 8.00 wt% CsCl and has a density of 1.0643 g/mL at 20°C. What is the boiling point of this solution? Kb = 0.51°C/m for water. Enter your answer with 2 decimal places and no units.

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:

To calculate the elevation in boiling point, we use the equation:

Or,

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:

Hence, the boiling point of solution is 100.53

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :