Question

How many grams of concentrated nitric acid solution should be used to prepare $250ml$ of $2.0M$ $HN{O_3}$ ? The concentrated acid is $70\% HN{O_3}$ .
(a) $45.0g$ conc. $HN{O_3}$
(b) $90.0g$ conc. $HN{O_3}$
(c) $70.0g$ conc. $HN{O_3}$
(d) $54.0g$ conc. $HN{O_3}$

Answer 1

In these questions, we have to know which formulas should be used to find how many grams of the acid should be used and what “The concentrated acid is $70\% HN{O_3}$ ” means and how we use this.
Formula used
Formula to calculate the moles of the solute present in the solution:
$n = M \times V$
Where, $n$ is the number of moles, $M$ is molarity of the solution and $V$ is the total volume of the solution in litres.
Weight by volume percentage formula:
$m = \dfrac{{100}}{{\% w/v}} \times n \times Mol$
Where, $\% w/v$ is the weight by volume percentage, $m$ is the weight of the solute in the grams, $n$ is the number of moles and $Mol$ is molecular mass.

Complete step-by-step answer
Starting from the given data:
Volume, $V = 250ml$
Molarity, $M = 2.0M$
Weight by volume percentage, $\% w/v = 70\%$
So, now starting to solve the question,
Here, the concentrated acid is $70\% HN{O_3}$ means that the $70$ grams of the concentrated acid is present in $100ml$ of the solution and the grams of the solute present in the solution of $250ml$ of $2.0M$ $HN{O_3}$ .
We have to prepare a solution of $250ml$ of $2.0M$ $HN{O_3}$ .
Given that the molarity of the solution is $2.0M$ that is $2$ moles of $HN{O_3}$ present in the $1$ litre of the solution.
So, the moles present in the $250ml$ of the solution,
$n = M \times V$
Putting the value of molarity and volume from the given data,
$n = M \times V = 2.0 \times \dfrac{{250}}{{1000}} = 0.5$
Hence, the number of moles of the present in the $250ml$ of the solution is $0.5$ .
The grams of $HN{O_3}$ present in the $70\% HN{O_3}$ required is,
$m = \dfrac{{100}}{{\% w/v}} \times n \times Mol$
Putting the value of moles, molecular mass and the percentage weight by volume,
$m = \dfrac{{100}}{{70}} \times 63 \times 0.5 = \dfrac{{100}}{{70}} \times 31.5 = 45g$
Hence, the grams of $HN{O_3}$ required is $45g$ .
Hence, the correct option is (A) $45.0g$ conc. $HN{O_3}$ .

Note
Weight by volume percentage is the weight of the solute present in per $100ml$ of the solution and molarity of the given solution states the number of moles of the solute present in per litre of the solution.