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Ways of Representing the Concentration of a Solution

0.7 g of Na...

Question

$0.7 g$ of $N a_{2} C O_{3} . x H_{2} O$ is dissolved in $100 m l, 20 m l$ of which required $19.8 m l$ of $0.1 N H C l$ . The value of $x$ is:

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Solution

The correct option is C $2$
$N a_{2} C O_{3} . X H_{2} O + 2 H C l \to 2 N a C l + (X + 1) H_{2} O + C O_{2}$

The strength of a solution is defined as the amount of solute in grams, present in one litre of solution. It is expressed in $g L^{- 1}$ . We are given that $0.7 g o f N a_{2} C O_{3} . X H_{2} O$ is dissolved 100 ml of solution.

The strength of the solution is therefore $= \frac{0.7}{[100 / 1000]} = 7 g L^{- 1}$

The molarity equation is written as $n 1 \times M 1 \times V 1 = n 2 \times M 2 \times V 2$ ....(1)
where n1-> acidity of $N a_{2} C O_{3}$
$M 1$ -> molarity of $N a_{2} C O_{3}$
$V 1$ -> volume of sodium carbonate solution used
$n 2$ -> basicity of $H C l$
$M 2$ -> molarity of $H C l$
$V 2$ -> volume of $H C l$ used

Since $H C l$ is monobasic, therefore its molarity is the same as its normality. Substituting the given values in equation (1), we get
$2 \times M 1 \times 20 = 1 \times 0.1 \times 19.8$

Therefore $M 1 = \frac{19.8}{400} = 0.0495 M$

Now $M o l a r i t y = \frac{S t r e n g t h i n g L^{- 1}}{m o l a r m a s s o f s o l u t e}$

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

Hence the value of $x$ here can be rounded off to $2$ .

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Similar questions

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0.7 g of Na2CO3.xH2O is dissolved in 100 ml,20 ml of which required 19.8 ml of 0.1 N HCl. The value of x is:

$0.7 g$ of $N a_{2} C O_{3} . x H_{2} O$ is dissolved in $100 m l, 20 m l$ of which required $19.8 m l$ of $0.1 N H C l$ . The value of $x$ is: