Determine the volume of diluted nitric acid $(d=1.11\; g\;ml^{-1}$, $19\%$ w/v) that can be prepared by diluting $50$ ml of conc. $HNO_{3}$ with water $(d= 1.42\; g\;ml^{-1}$, $76\%$ w/v).

Using the mass-volume relationship for dilution, the initial amount of solute is 50 ml multiplied by 0.76 g/ml, giving 38 g of HNO3. The final volume of the 19% w/v solution is the total mass of solute divided by the final concentration (0.19 g/ml), which equals 200 ml. The result is 200 ml.

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