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

Hint: 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.