Question

Calculated Molarity of following solutions

24.5g of sodium nitrate dissolved in a 275ml solution.

25.0 mL of a 0.125 M hyrochloric acid solution diluted with water to 500.0 mL.

A solution of nitric acid with pH= 2.42

Answer 1

(i) 24.5g sodium nitrate dissolved in 275ml solution.

Moar mass of NaNO₃ = (1 x 23) + (1 x 14) + (3 x 16)

= 85g/mol

Mass of NaNO3 dissolved = 24.5g

Number of moles of NaNO3 dissolved = mass/ molar mass

= 24.5g/85g per mol.

= 0.288 mol

Molarity = Number of moles of solute/Volume of solution in Litres

= 0.288 mol/0.275 L

[Volume = 275ml = 0.275L]

= 1.048 mol L^-1

(ii) 25ml of 0.125, hydrochloric acid solution diluted with water to 500ml.

- First we will calculate the number of moles of HCl present in solution and then the new molarity.