Question

Saltpetre is a rock which occurs in Chile. It contains iodine in the form of

sodium iodate(V), NalO,.
Sodium thiosulphate solution was used to find the concentration of a solution of
sodium iodate(V).
(i) The experiment used 36-0 cm' of a sodium thiosulphate solution of
concentration 0-100 mol dm.
Calculate the number of moles of sodium thiosulphate present in this solution.
(ii) Six moles of sodium thiosulphate were needed for each mole of the
sodium iodate(V).
State the number of moles of sodium iodate(V) present in the sample.
[1]
(iii) The volume of sodium iodate(V) solution used in the experiment was 50-0cm'.
Calculate the concentration of the sodium iodate(V) solution in mol dm. [1]
(iv) Use your answer to (iii) to calculate the concentration of the sodium iodate(V)
solution in g dm.
[1]
The relative molecular mass of sodium iodate(V) is 198.

Answer 1

(i) To calculate the number of moles of sodium thiosulphate present in the solution, we can use the formula:

moles = concentration x volume

The concentration of the sodium thiosulphate solution is given as 0.100 mol dm^-3 and the volume is 36.0 cm^3.

We can calculate the number of moles of sodium thiosulphate as:

moles = 0.100 mol dm^-3 x (36.0 cm^3 / 1000 cm^3/dm^3) = 0.0036 mol

(ii) We are told that six moles of sodium thiosulphate were needed for each mole of sodium iodate(V). So, we can calculate the number of moles of sodium iodate(V) in the sample as:

moles of sodium iodate(V) = moles of sodium thiosulphate / 6

moles of sodium iodate(V) = 0.0036 mol / 6 = 6.00 x 10^-4 mol

(iii) To calculate the concentration of the sodium iodate(V) solution in mol dm^-3, we can use the formula:

concentration = moles of solute / volume of solution

The number of moles of sodium iodate(V) is 6.00 x 10^-4 mol and the volume of the solution is 50.0 cm^3.

so,

concentration = 6.00 x 10^-4 mol / (50.0 cm^3 / 1000 cm^3/dm^3) = 1.2 x 10^-4 mol dm^-3

(iv) To calculate the concentration of the sodium iodate(V) solution in g dm^-3, we can use the formula:

concentration in g dm^-3 = molar mass x concentration in mol dm^-3

The molar mass of sodium iodate(V) is 198 g/mol.

so,

concentration in g dm^-3 = 198 g/mol x 1.2 x 10^-4 mol dm^-3 = 0.00237 g dm^-3

This is the final concentration of sodium iodate(V) solution in g dm^-3.