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# Determine the volume of diluted nitric acid (d = 1.11 g/ml, 19 % w/v) that can be prepared by diluting 50 ml of conc. $HN{O_3}$ with water (d = 1.42 g/ml, 76 % w/v).a.) 200 mlb.) 300 mlc.) 400 mld.) 500 ml

Last updated date: 19th Jun 2024
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Hint: The word w/v means weight by volume. We have a percentage of weight by volume solutions. From that, we can find the amount of nitric acid. Then, we can find the volume after dilution from that amount.

Complete Solution :
We have been given in question that the solution to be taken is 76 % w/v.
So, the 50 ml of 76 % w/v solution has = $50 \times \dfrac{{76}}{{100}}$
50 ml of 76 % w/v solution has = 38 g of nitric acid
If we suppose ‘V’ ml of solution is formed after dilution.
So, ‘V’ ml of 19 % w/v solution has = $V \times \dfrac{{19}}{{100}}$ = 38 g of nitric acid
V = $\dfrac{{38 \times 100}}{{19}}$
V = 200 ml
So, we can have 200 ml of new solution by using 50 ml of first solution.
So, the correct answer is “Option A”.

Note: It must be noted that the solution is diluted. This means the water is added to the solution. We can observe from densities that the first solution has a density of 1.42 g/ml. This density has been decreased to 1.11 g/ml in the second solution. So, the new solution will have 50 ml of the first nitric acid solution and 150 ml of water.