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Ways of Representing the Concentration of a Solution

0.70g of a sa...

Question

$0.70 g$ of a sample of $N a_{2} C o_{3} . x H_{2} O$ were dissolved in water and the volume was made to $100 m l; 20 m l$ of this solution required $19.8 m l$ of $N / 10 H C l$ for complete neutralisation.the value of $x$

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Solution

The correct option is C 2
Reaction:

$N a_{2} C O_{3} . x H_{2} O + 2 H C l \to 2 N a C l + (x + 1) H_{2} O + C O_{2}$

The strength of a solution is defined as the amount of solute in grams, present in one litre of solution. It is expressed in $g L^{-}$ .
We are given that $0.7 g$ of $N a_{2} C O_{3} . x H_{2} O$ is dissolved in $100$ ml of solution.

The strength of the solution is therefore = $\frac{0.7}{[100 / 1000]} = 7 g L^{-}$

The molarity equation is written as

$n_{1} \times M_{1} \times V_{1} = n_{2} \times M_{2} \times V_{2}$ ....(1)

where
$n_{1} \to$ acidity of $N a_{2} C O_{3}$
$M_{1} \to$ molarity of $N a_{2} C O_{3}$
$V_{1} \to$ volume of sodium carbonate solution used
$n_{2} \to$ basicity of $H C l$
$M_{2} \to$ molarity of $H C l$
$V_{2} \to$ volume of $H C l$ used

Since $H C l$ is monobasic, therefore its molarity is the same as its normality\basicity. Substituting the given values in equation (1), we get

$2 \times M_{1} \times 20 = 1 \times 0.1 \times 19.8$

$∴ M_{1} = \frac{19.8}{400} = 0.0495 M$

Now $M o l a r i t y = \frac{strength}{molar mass of solute}$

Therefore, molar mass of $N a_{2} C O_{3} . x H_{2} O = \frac{7}{0.0495} = 141.414 g / m o l$

But molar mass of $N a_{2} C O_{3} . x H_{2} O$ = Mass of anhydrous $N a_{2} C O_{3}$ + mass of $x$ molecules of water $= 106 + 18 x$

Therefore $106 + 18 x = 141.414$ , which gives $x =$ $\frac{35.414}{18} = 1.976$

Hence the value of $x$ here can be rounded off to $2.$ Hence, option C is correct.

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Similar questions

0.7 g of Na₂Co₃.XH₂O were dissolved in water and the volume was made to 100mL , 20 mL of this solution required 19.8 mL of N/10 HCl for complete neutralisation, the value of 'x' is

A) 7

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0.70g of a sample of Na2Co3.xH2O were dissolved in water and the volume was made to 100ml;20ml of this solution required 19.8ml of N/10HCl for complete neutralisation.the value of x​

$0.70 g$ of a sample of $N a_{2} C o_{3} . x H_{2} O$ were dissolved in water and the volume was made to $100 m l; 20 m l$ of this solution required $19.8 m l$ of $N / 10 H C l$ for complete neutralisation.the value of $x$