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Ways to Define Concentration

250mL of sodi...

Question

$250 m L$ of sodium carbonate solution contains $2.65 g$ of ${N a}_{2} {C O}_{3}$ . If $10 m L$ of this solution is diluted to $500 m L$ , the concentration of the diluted acid will be?

0.01 M

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0.001 M

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0.05 M

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0.002 M

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Solution

The correct option is C $0.002 M$
Molarity = $\frac{n \times 1000}{V_{m l}}$

Molar mass of $N a_{2} C O_{3} = 106 g / m o l$

No of moles of $N a_{2} C O_{3} = \frac{2.65}{106} = 0.025 m o l$

Molarity = $\frac{0.025 \times 1000}{250} = 0.1 M$
Dilution Formula

$M_{1} V_{1} = M_{2} V_{2}$

$M_{2} = \frac{M_{1} \times V_{1}}{V_{2}}$

$M_{2} = \frac{0.1 \times 10}{500} = 0.002 M$

Option D is correct.

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Similar questions

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