Question

Calculate the molalities of some commerical reagents from thefollowing data:

HCl has a formula weight (amu) of 36.465, Density of thesolution(g/mL) of 1.19, Weight % of 37.2, and Molarity of12.1.
HC2H3O2 has a formula weightof 60.05, Density of 1.05, Weight % of 99.8, and Molarity of17.4.
NH3(aq) has a formula weight of 17.03, Denisty of0.90, Weight % of 28.0, and Molarity of 14.8

Answer 1

The molality of HCl solution is 16.24 mol/kg.

The molality of
`HC_2H_3O_2` solution is 82,500 mol/kg.

The molality of
`NH_3` solution is 27.78 mol/kg.

Step-by-step explanation:

formula used:

`Molality=\frac{Moles}{\text{Mass of solvent(kg)}}`

1) Mass percentage of the HCl solution = 37.2%

This means that in 100 grams of solution 37.2 grams of HCl is present.

Mass of HCl (solute)= 37.2 g

Mass of water(solvent) = 100 g - 37.2 g = 62.8 g = 0.0628 kg (1g = 0.001 kg)

Mole of HCl =
`(37.2 g)/(36.465 g/mol)=1.020 mol`

`Molality=\frac{Moles}{\text{Mass of solvent(kg)}}`

`m=(1.020 mol)/(0.0628 kg)=16.24 mol/kg`

The molality of HCl solution is 16.24 mol/kg.

2) Mass percentage of the
`HC_2H_3O_2` solution = 99,8%

This means that in 100 grams of solution 99.8 grams of
`HC_2H_3O_2` is present.

Mass of
`HC_2H_3O_2`(solute)= 99.8 g

Mass of water(solvent) = 100 g - 99.8 g = 0.2 g = 0.0002 kg (1g = 0.001 kg)

Mole of
`HC_2H_3O_2` =
`(99.8 g)/(60.05g/mol)=16.50 mol`

`Molality=\frac{Moles}{\text{Mass of solvent(kg)}}`

`m=(16.50 mol)/(0.0002 kg)=82,500 mol/kg`

The molality of
`HC_2H_3O_2` solution is 82,500 mol/kg.

3) Mass percentage of the
`NH_3` solution = 28.0%

This means that in 100 grams of solution 28.0 grams of
`NH_3` is present.

Mass of
`NH_3`(solute)= 37.2 g

Mass of water(solvent) = 100 g - 28.0 g = 72.0 g = 0.072 kg (1g = 0.001 kg)

Mole of
`NH_3`=
`(28.0g)/(17 g/mol)=2mol`

`Molality=\frac{Moles}{\text{Mass of solvent(kg)}}`

`m=(2 mol)/(0.072kg)=27.78 mol/kg`

The molality of
`NH_3` solution is 27.78 mol/kg.