Plz me with this problem what is the boiling point (in celsius) of a 1.56m aqueous solution of cacl2?

The answer is 102.39* C

Hope this helps

102.4 °C

Explanation:

There are four colligative properties :

a) elevation in boiling point

b) depression in freezing point

c) relative lowering of vapor pressure

d) osmotic pressure

When we add a non volatile solute in a solvent, the solution has higher boiling point as compared to the pure solvent, this is known as elevation in boiling point.

The elevation in boiling point is related to molality as:

elevation in boiling point =i X KbXmolality

Kb (water) = 0.512

molality given as = 1.56 m

i = vant hoff factor =3 (for calcium chloride)

Putting values

elevation in boiling point = 3X0.512X1.56 = 2.4

Thus the boiling point of aqueous solution will be = 100+2.4 = 102.4 °C

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 :