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Standard XII

Chemistry

Stoichiometry

0.7 gm of n...

Question

0.7 gm of $n A_{2} C O_{3} . x H_{2} O$ is dissolved in 100 ml, 20 ml of which required 19.8 ml of 0.1 N HCl. The value of x is :

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Solution

The correct option is C 2
$\frac{0.7}{100} \times 1000 = 7 g m / L$
$2 \times M_{1} \times 20 = 1 \times 0.1 \times 19.8$
$M_{1} = 0.0495$
$m o l a r i t y = \frac{s t r e n g t h}{m o l e c u l a r m a s s}$
$M . M = \frac{7}{0.0495} = 141.4 g$
$x = 2$

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Similar questions

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:

0.7 g

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

\frac{N}{10} H C l

for completing neutralisation. The value of

x

is:

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

1.20 g

sample of

N a_{2} C O_{3}

and

K_{2} C O_{3}

was dissolved in water to form

100 m L

of a solution.

20 m L

of this solution required

40 m L

0.1 N H C l

for complete neutralization. Calculate the weight of precipitate in the mixture if another

20 m L

of this solution is treated with excess of

B a C l_{2}

Q. A 20 ml solution of 0.1 N

N a_{2} C O_{3}

is titrated with 0.1N HCl with phenolphthalein as the indicator. Another 20 ml solution of 0.1 N

N a_{2} C O_{3}

is titrated with 0.1 N HCl with methyl orange as the indicator. The volume of HCl required for both the titrations resepectively is:

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0.7 gm 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 :

The correct option is C 20.7100×1000=7gm/L2×M1×20=1×0.1×19.8M1=0.0495molarity=strengthmolecularmassM.M=70.0495=141.4gx=2

0.7 gm of $n A_{2} C O_{3} . x H_{2} O$ is dissolved in 100 ml, 20 ml of which required 19.8 ml of 0.1 N HCl. The value of x is :

The correct option is C 2
$\frac{0.7}{100} \times 1000 = 7 g m / L$
$2 \times M_{1} \times 20 = 1 \times 0.1 \times 19.8$
$M_{1} = 0.0495$
$m o l a r i t y = \frac{s t r e n g t h}{m o l e c u l a r m a s s}$
$M . M = \frac{7}{0.0495} = 141.4 g$
$x = 2$