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The molarity of resultant solution when 100 ml of 0.3 M $$\text{HNO}_3$$ and 200 ml of 0.3 M $$\text{H}_2\text{SO}_4$$ solution mixed together
Moles of acid are $$0.1 \times 0.3 = 0.03$$ of $$\text{HNO}_3$$ and $$0.2 \times 0.3 = 0.06$$ of $$\text{H}_2\text{SO}_4$$, a total of 0.09 mol in 0.3 L. The molarity of the mixture is $$\frac{0.09}{0.3} = 0.3\text{ M}$$ (both solutions were 0.3 M, so mixing does not change it).
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