Question

10 gm of a hydrated sodium carbonate $N{{a}_{2}}C{{O}_{3}}.x{{H}_{2}}O$ on strong heating loses a weight of 6.3 gm. The value of x is
(A) 5
(B) 7
(C) 1
(D) 10

Answer 1

Hint: Sodium carbonate on heating loses the water molecule called the water of crystallisation. Some weight is lost in it. The moles of water lost need to be found to get the value of x.

Complete step by step answer:
- In lower classes of physical chemistry, we have dealt with the topics of calculation of number of moles in a substance and also some related calculations. Let us see the calculation of the number of moles of water in the given question.
- We have been given the mass of sodium carbonate that is 10 g
- So mass of $N{{a}_{2}}C{{O}_{3}}.x{{H}_{2}}O$ that is ${{w}_{Hy}}$= 10 g
- Mass of water produced,${{w}_{{{H}_{2}}O}}$ is given as 6.3 g
- Therefore, the molar mass of hydrated sodium carbonate can be calculated as$106+18x$g/mol
Now, we know that the molar mass of water molecule is 18 g/mol
Therefore, the number of moles of water will be,${{n}_{{{H}_{2}}O}}=\dfrac{6.3}{18}=0.35$ moles
-Now number of moles of hydrated sodium carbonate can be thus identified similarly as, ${{n}_{Hy}}=\dfrac{10}{106+18x}$ moles.
we have to now find the value of x and that will be equal to the ratio of number of moles of water to that of number of moles of hydrated sodium carbonate that is $\dfrac{{{n}_{{{H}_{2}}O}}}{{{n}_{Hy}}}$
-Now moles of$N{{a}_{2}}C{{O}_{3}}.x{{H}_{2}}O$ = number of moles of $N{{a}_{2}}C{{O}_{3}}$
- Also, mass of sodium carbonate,${{w}_{N{{a}_{2}}C{{O}_{3}}}}={{w}_{Hy}}-{{w}_{{{H}_{2}}O}}$
This will be equal to $10 - 6.3 = 3.7$g
- Moles of sodium carbonate, $N{{a}_{2}}C{{O}_{3}}=\dfrac{{{w}_{N{{a}_{2}}C{{O}_{3}}}}}{{{M}_{N{{a}_{2}}C{{O}_{3}}}}}=\dfrac{3.7}{106}=0.0349$ moles
- Solving the equation for x we get,
\[x=\dfrac{{{n}_{{{H}_{2}}O}}}{{{n}_{N{{a}_{2}}C{{O}_{3}}}}}=\dfrac{0.35}{0.0349}\approx 10\]
- Therefore the value of x is ten. The correct answer is option “D” .

Note: Sodium carbonate is obtained as three hydrates and anhydrous salt. Sodium carbonate decahydrate readily efflorescence to form monohydrate. The mass of hydrated salt and dehydrated salt are equal.