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Molality

When 25 mL ...

Question

When $25$ mL of $0.1 M$ $A g N O_{3}$ is mixed with $50.0$ mL of $0.08 M$ $N a_{2} C O_{3}$ solution, then the molar concentration of $N a_{2} C O_{3}$ in the solution is $x \times 10^{- 2} M$ . Here $x$ is:

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Solution

The number of moles of unreacted sodium carbonate $= M_{2} V_{2} - M_{1} V_{1} = 0.08 \times 0.05 - 0.1 \times 0.0025 = 0.004 - 0.0025 = 0.0015 M$ .
Total volume $= 75 m L = 0.075 L$ .
The molar concentration of sodium carbonate is $\frac{0.0015 M}{0.075 L} = 0.02 M = 2 \times 10^{- 2} M$ .

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