You visited us 1 times! Enjoying our articles? Unlock Full Access!

Mole Concept and Avogadro's Number

500 mL of 2...

Question

$500 m L$ of $2 M H C l . 100 m L$ of $2 M H_{2} S O_{4}$ and one g equivalent of a monoacidic alkali are mixed together. $30 m L$ of this solution required $20 m L$ of $N a_{2} C O_{3} . x H_{2} O$ solution obtained by dissolving $143 g N a_{2} C O_{3} . x H_{2} O$ in one litre solution. Calculate water of crystallisation of $N a_{2} C O_{3} . x H_{2} O .$

Open in App

Solution

500 ml of 2M HCl +100ml of 2M $H_{2} S O_{4}$

Total milliequivalents of Hydronium ion will be $\to$ 500 x 2 + 100 x 4{since H2SO4 will give 2 H+} = 1400milliequivalents = 1.4 equivalents

The equivalents of $O H^{-}$ added in form of Monobasic alkali=1

Net after neutralization will be 0.4 equivalents of Hydronium ion.

30ml of this solution required 20ml of 143gm $N a_{2} C O_{3} . x H_{2} O$ in one-litre solution.

Now just equate both milliequivalents.

We know,

$2 H^{+} + N a_{2} C O_{3} \to N a O H + C O_{2}$

THAT MEANS 2 HYDRONIUM IONS ARE NEEDED FOR EACH $N a_{2} C O_{3}$ MOLECULE

n factor for $N a_{2} C O_{3}$ is 2.

Molarity of $N a_{2} C O_{3}$ = $\frac{143}{106 + 18 x}$

Normality = M x n factor =2M

$0.4 \times 30 m L = 2 \times \frac{143}{106 + 18 x} \times 20$

$∴$ x = 10

Suggest Corrections

Similar questions

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

N / 10 H C l

for complete neutralisation.the value of

x

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

B) 3

C) 2

D) 5

0.7 g

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

0.1 N H C l

. The value of

x

is:

14.3

g of

N a_{2} C O_{3} . x H_{2} O

completely neutralises

100

ml of

1 N H_{2} S O_{4}

solutions, Value of x is

5 m L

8 N H N O_{3}, 4.8 m L

5 N H C l

and a certain volume of

17 M H_{2} S O_{4}

are mixed together and made upto

2 L

30 m L

of this acid mixture exactly neutralizes

42.9 m L

{N a}_{2} C O_{3}

solution containing

1 g

{N a}_{2} C O_{3}

10 H_{2} O

100 m L

of water. The mass (in

g

) of sulphate ions present in the solution is (only integer part) :

Join BYJU'S Learning Program

500mL of 2MHCl.100mL of 2MH2SO4 and one g equivalent of a monoacidic alkali are mixed together. 30mL of this solution required 20mL of Na2CO3.xH2O solution obtained by dissolving 143gNa2CO3.xH2O in one litre solution. Calculate water of crystallisation of Na2CO3.xH2O.