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Standard XII

Chemistry

Molarity

250 ml of N...

Question

250 ml of ${N a}_{2} {C O}_{3}$ solution contains 2.65 g of ${N a}_{2} {C O}_{3}$ .10 ml of this solution is mixed with $X$ ml of water to obtain 0.001 M ${N a}_{2} {C O}_{3}$ solution. The value of $X$ is:

1000

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990

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9990

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Solution

The correct option is B 990
The molar mass of sodium carbonate is 106 g/mol.
2.65 g of sodium carbonate corresponds to
$\frac{2.65}{106} = 0.025$ moles
Molar concentration will be the ratio of the number of moles of sodium carbonate to the volume of solution. It will be
$\frac{0.025}{0.250} = 0.1$ M.
0.1 M solution is diluted to 0.001 M
$M V = M^{'} V^{'}$
$0.1 \times 10 = 0.001 \times (10 + X)$
$10 + X = 1000$
$X = 1000 - 10 = 990 m l$
Hence, the value of $X$ is $90$ .

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Similar questions

250 m L

of a

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

solution contains

2.65 g

{N a}_{2} {C O}_{3} \cdot 10 m L

of this solution is added to

x

m L

of water to obtain

0.001 M

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

solution. The value of

x

is:

[Molecular weight of

{N a}_{2} {C O}_{3} = 106

]

250 m l

of a sodium carbonate solution contains

2.65

grams of

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

10 m l

of this solution is added to

^{'} x^{'} m l

of water to obtain

0.001 M

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

solution.

What is the value of

^{'} x^{'}

m l

[Molecular weight of

{N a}_{2} {C O}_{3} = 106]

Q. Solution "X" contains

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

and

N a H C O_{3}

. 20 ml of X when titrated using methyl orange indicator consumed 60 ml of 0.1 M

H C l

solution. In another experiment , 20 ml of X solution when titarted using phenolphthalein consumed 20 ml of 0.1 M

H C l

solution. The concentrations (in

m o l {l i t}^{- 1}

) of

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

and

N a H C O_{3}

in X are respectively :

Q. 250 ml of sodium carbonate solution contains 2.65g of

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

. If 10 ml of the solution is diluted to 1 L. What is the concentration of the resultant solution?

Q. A sample of

A g C l

was treated with

5

mL of

1.5 M

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

solution to give

{A g}_{2} {C O}_{3}

. The remaining solution contained

0.0026 g

{C l}^{-}

per litre. The solubility product of

A g C l

(

K_{s p}

_{({A g}_{2} {C O}_{3})} = 8.2 \times 10^{- 12}

)

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250 ml of Na2CO3 solution contains 2.65 g of Na2CO3 .10 ml of this solution is mixed with X ml of water to obtain 0.001 M Na2CO3 solution. The value of X is:

250 ml of ${N a}_{2} {C O}_{3}$ solution contains 2.65 g of ${N a}_{2} {C O}_{3}$ .10 ml of this solution is mixed with $X$ ml of water to obtain 0.001 M ${N a}_{2} {C O}_{3}$ solution. The value of $X$ is: