Question

The hydrated salt $N{a_2}S{O_4}.10{H_2}O$ undergoes X% loss in weight on heating and becomes anhydrous. The value of X will be:
A. 10
B. 45
C. 56
D. 70

Answer 1

On heating $N{a_2}S{O_4}.10{H_2}O$ becomes anhydrous that implies that ${H_2}O$ molecules get separated. So write the equation by keeping this in mind. Balance the chemical equation before you proceed with the answer.

Complete step by step solution:
According to the question water molecules gets removed so balanced chemical equation becomes:
$N{a_2}S{O_4}.10{H_2}O \to N{a_2}S{O_4} + 10{H_2}O$
No. of moles of $N{a_2}S{O_4}.10{H_2}O$ in given equation= 1 mole
No. of moles of $N{a_2}S{O_4}$ in given equation= 1 mole
No. of moles of ${H_2}O$ in given equation= 10 moles
Weight of substance = No. of moles of substance$\times$Molecular mass of substance
Therefore we need to find out Molecular Mass of $N{a_2}S{O_4}.10{H_2}O$
Molecular mass of $N{a_2}S{O_4}.10{H_2}O$ = $23 \times 2 + 32 + 16 \times 4 + 10(2 \times 1 + 16)$
$= \,46+32+64+180= 322\, g/mole$
Hence Weight of $N{a_2}S{O_4}.10{H_2}O$ = No. of moles of N{a_2}S{O_4}.10{H_2}O$$$ \times $Molecular mass of N{a_2}S{O_4}.10{H_2}O$Therefore, Weight of$N{a_2}S{O_4}.10{H_2}O= 1,mol \times 322,g/mol$$
$=322\,g$
Similarly Molecular Mass of $N{a_2}S{O_4} = 23 \times 2+32+16 \times 4$
$= \,142\,g/mole$
And Molecular mass of water=$2 \times 1+16$
$=\,18\,g/mole$
So weight of water= No. of moles of water$\times$Molecular mass of water
$=18 \times 10= 180\,g$
And weight of $N{a_2}S{O_4}$ = No. of moles of $N{a_2}S{O_4}$ $\times$Molecular mass of $N{a_2}S{O_4}$
$= 142 \times 1= 142\,g$
$\therefore Percentage\, of\, loss\, in\, weight= \dfrac{(Initial\, weight-Final\, weight)}{Total \,Weight} \times 100\\\ =\dfrac{(322-142)}{322} \times 100\\\ =\dfrac{18000}{322}\\\ =55.9\%$
This is approximately equal to $56\%$

**Therefore the correct answer to the question is option (3).

Note:**
1. Percentage loss of weight of a substance is equal to = (Initial weight-Final weight) 100/(Total Weight).
2. Balance the equation of reaction before solving the question. Otherwise you will not be able to find out the correct no. of moles of substance which will lead to incorrect answers.