0-7 g of Na2CO3- xH2O was dissolved in water and the volume made up to 100 mL- 20 mL of it required 19-8 mL of N10HCl for completing neutralisation- The value of x is-

The correct option is C 2 Number of milliequivalents of Na_2CO_3· xH_2O in 20 mL #160; = 19.8×110 #160; #160; #160; #160; ...

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Byju's Answer

Standard XII

Chemistry

Stoichiometric Calculations

0.7 g of Na...

Question

$0.7 g$ of $N a_{2} C O_{3} \cdot x H_{2} O$ was dissolved in water and the volume made up to $100 m L, 20 m L$ of it required $19.8 m L$ of $\frac{N}{10} H C l$ for completing neutralisation. The value of $x$ is:

7

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Solution

The correct option is C $2$
Number of milliequivalents of $N a_{2} C O_{3} \cdot x H_{2} O$ in $20 m L$ $= 19.8 \times \frac{1}{10}$
$= 1.98$

$∴$ Number of milliequivalents in $100 m L = 1.98 \times 5 = 9.9$

$\frac{M a s s}{E q u i v a l e n t m a s s} \times 1000 = 9.9$

$\frac{0.7}{M / 2} \times 1000 = 9.9$

$M = 141.40$

(M is the molar mass of $N a_{2} C O_{3} \cdot x H_{2} O$ )

$106 + 18 x = 141.40$

$x = 2$ .

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0.7 g of Na2CO3⋅xH2O was dissolved in water and the volume made up to 100 mL,20 mL of it required 19.8 mL of N10HCl for completing neutralisation. The value of x is:

The correct option is C 2Number of milliequivalents of Na2CO3⋅xH2O in 20 mL =19.8×110 =1.98∴ Number of milliequivalents in 100 mL=1.98×5=9.9MassEquivalent mass×1000=9.90.7M/2×1000=9.9M=141.40(M is the molar mass of Na2CO3⋅xH2O)106+18x=141.40x=2.

$0.7 g$ of $N a_{2} C O_{3} \cdot x H_{2} O$ was dissolved in water and the volume made up to $100 m L, 20 m L$ of it required $19.8 m L$ of $\frac{N}{10} H C l$ for completing neutralisation. The value of $x$ is: