Question

0.7g of Na2Co3.xH2O is dissolved in 100ml ,20ml of which required19.8ml of 0.1N Hcl. The value of x is :

Answer 1

The equation given in the question is;

$Na_2CO_3.XH_2O +2HCl$ → $2NaCl + (X + 1)H_2O +CO_2$

We know that, the strength of a solution can be measured by the amount of solute (in grams) present in one litre of solution, expressed in terms of $gL^-^1$

According to the question,

0.7g of $Na_2CO_3.XH_2O$ has been dissolved in a solution of 100ml.

∴ Strength of the solution = $\frac{0.7}{\frac{100}{1000} } = 7 gL^-^1$

Now, we know that, molarity = $n1$ × $M1$ × $V1$ = $n2$ × $M2$ × $V2$ ...(Eqn. 1)

in which,

$n1$ = acidity in $Na_2CO_3$

$M1$ = molarity in  $Na_2CO_3$

$V1$ = volume of  $Na_2CO_3$ present

$n2$ = basicity in $HCl$

$M2$ = molarity of $HCl$

$V2$ = volume of $HCl$ present

∵  $HCl$ is for being monobasic in nature, its molarity = its normality.

On substituting the given values in Eqn. 1, we get;

2 × $M1$ × 20 = 1 × 0.1 × 19.8

⇒ $M1$ = $\frac{19.8}{400}$ M = 0.0495 M

Then, Molarity = (strength in $gL^-^1$) ÷ molar mass of solute

∴ Molar mass of  $Na_2CO_3.XH_2O$  = $\frac{7}{0.0495} g/mol = 141.414 g/mol$

However, the molar mass of $Na_2CO_3.XH_2O$ = (Mass of anhydrous $Na_2CO_3$) + mass of $X$ molecules of water = 106 + 18$X$

∴ 106 + 18$X$ = 141.414

⇒ $X$ = $\frac{35.414}{18}$ = [1.976] = 2

Ans) Value of $X$ = 2