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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 f...

Glauber’s salt is Na2SO4.nH2ON{a_2}S{O_4}.n{H_2}O , 16.1g16.1g of Glauber’s salt is heated to remove water from Atkin completely. Mass of the remaining solid sample is 7.1g7.1g . What is the value of nn ?

Explanation

Solution

Given salt is Glauber’s salt is Na2SO4.nH2ON{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 16.1g16.1g 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 7.1g7.1g
The mass of solid sodium sulphate is 7.1g7.1g where the molar mass is 142gmol1142gmo{l^{ - 1}}
Thus, the moles of sodium sulphate (Na2SO4)\left( {N{a_2}S{O_4}} \right) is 7.1g142gmol1=0.05moles\dfrac{{7.1g}}{{142gmo{l^{ - 1}}}} = 0.05moles
The mass of water will be (16.107.10)g\left( {16.10 - 7.10} \right)g and the molar mass of water is 18gmol118gmo{l^{ - 1}}
The moles of water will be (16.107.10)g18gmol1=0.5mol\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
nNa2SO4:nH2O=0.050.05:0.50.05{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 1:101:10
Thus, the molecular formula of the hydrated salt will be Na2SO4.10H2ON{a_2}S{O_4}.10{H_2}O
Thus, the value of n is 1010

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.