Question

19. Calculate the molality of Ca(OH)2 in a 1.50 M aqueous solution that has a density of 1.320 g/mL.

Answer 1

Answer:

[a] 30.0 g of Ca is added to 30.0 g of water and allowed to react according to the

following equation:

Ca(s) + 2H2O(l) + Ca(OH)2(aq) + H2(g)

[i] Calculate the amount in mole of each reactant.

[ii] Which reactant is the limiting reagent?

[iii] What is the mass of the Ca(OH)2 formed?

[b] Calculate the energy and frequency of radiation emitted when an electron in

hydrogen atom returns from 5th energy level to 2nd energy level.

Answer 2

Answer:

we know that in 1 L of solution there are 1.5 mol of Ca(OH)↓2

let's figure out mass of the solution

Mass of solution = 1L*1000mL/1L*1.320g/1 ml

=1320gm of solution.

Mass of water= mass of solution - mass of Ca(OH)↓2 in the solution

=1320g - [1.5 mol of Ca(OH)↓2 * 74.093g/ 1 mol Ca(OH)↓2]

=1320g - 111.13g of Ca(OH)↓2

=1208g of water.

=1.208 kg of water.

Now use the units of molality as an equation,

Molality (Mol solute/kg solvent) = no. mol Ca(OH)↓2/ no. kg water

so,

Molality= 1.5mol of Ca(OH)↓2/ 1.208kg of water

Molality= 1.24m of Ca(OH)↓2