Question

Glauber’s salt is \[N{a_2}S{O_4}.n{H_2}O\] , \[16.1g\] of Glauber’s salt is heated to remove water from Atkin completely. Mass of the remaining solid sample is \[7.1g\] . What is the value of \[n\] ?

Answer 1

Hint: Given salt is Glauber’s salt is \[N{a_2}S{O_4}.n{H_2}O\] it is a hydrated salt consisting of n number of moles of water. The number of moles of sodium sulphate and water can be determined from the ratio of mass to molar mass. By combining the mole ratio, the value of n can be determined.

Complete answer:
Given that \[16.1g\] of Glauber’s salt is heated to remove water from crystals. By heating a hydrated salt, the water molecules can be evaporated, which leads to the formation of a dehydrated salt.
Given that the mass of remaining solid sample is \[7.1g\]
The mass of solid sodium sulphate is \[7.1g\] where the molar mass is \[142gmo{l^{ - 1}}\]
Thus, the moles of sodium sulphate \[\left( {N{a_2}S{O_4}} \right)\] is \[\dfrac{{7.1g}}{{142gmo{l^{ - 1}}}} = 0.05moles\]
The mass of water will be \[\left( {16.10 - 7.10} \right)g\] and the molar mass of water is \[18gmo{l^{ - 1}}\]
The moles of water will be \[\dfrac{{\left( {16.10 - 7.10} \right)g}}{{18gmo{l^{ - 1}}}} = 0.5mol\]
Divide the moles of Sodium sulphate and water with moles of sodium sulphate to obtain the mole ratio
\[{n_{N{a_2}S{O_4}}}:{n_{{H_2}O}} = \dfrac{{0.05}}{{0.05}}:\dfrac{{0.5}}{{0.05}}\]
The simplest mole ratio will be \[1:10\]
Thus, the molecular formula of the hydrated salt will be \[N{a_2}S{O_4}.10{H_2}O\]
Thus, the value of n is \[10\]

Note:
The molar mass of sodium sulphate can be exactly taken and the mass will be the mass of dehydrated salt. The mass of water can be obtained by subtracting the mass of dehydrated salt from the mass of hydrated salt gives the mass of water molecules.