Question

How many grams of concentrated nitric acid solution should be used to prepare 250mL of 2 M $$HN{O}_3$$ ? The concentrated acid is 70 % (w/w) $$HN{O}_3$$.
A. 90.0g conc. $$HN{O}_3$$
B. 70.0g conc. $$HN{O}_3$$
C. 54.0g conc. $$HN{O}_3$$
D. 45.0g conc. $$HN{O}_3$$

Answer 1

Hint: Molarity is the number of moles of solute per liter of solution. That means if we can calculate the amount of solute in 1000mL of solution, we can begin to proceed towards our answer.

Complete step by step solution:
We know,
2M $$HN{O}_3$$ means 2 moles of $$HN{O}_3$$ in 1000mL of solution.
Therefore, in 250mL solution we have $${\displaystyle \frac{2}{1000}}\times 250=0.5moles$$
Molecular weight of $$HN{O}_3$$ = (1x1) + (1x14) + (3x16) = 63 grams.

We now calculate the weight of $$HN{O}_3$$ in 250 mL of solution.
We know,
$$weightofsolute=no.ofmolesofsolute\times molecularmass$$
Substituting the values in this equation we get,
Weight of $$HN{O}_3$$ = 0.5 x 63g
70% weight by weight means 70g $$HN{O}_3$$
 is present in 100g of solution.
Therefore, according to our calculation,
The mass of $$HN{O}_3$$ required is $${\displaystyle \frac{100}{70}}\times 0.5\times 63=45grams$$
Hence, Option (D) 45.0g of conc. $$HN{O}_3$$ is the correct answer.

Note: Instead of calculating the moles of solute and liters of solution present individually, we can also string all the calculations together in one problem.
To calculate the molarity of a solution, it is essential to remember the number of moles of solute must be divided by the total liters of solution produced. If the amount of solute is given in grams, we must first calculate the number of moles of solute using the solute’s molar mass, then calculate the molarity using the number of moles and total volume.