Question

How many grams of concentrated nitric acid solution should be used to prepare $250$ mL of $2$M $HN{O}_3$? The concentrated acid is $70\%$(w/w) $HN{O}_3$.

• A. $90.0$ g conc. $HN{O}_3$
• B. $70.0$g conc. $HN{O}_3$
• C. $54.0$g conc. $HN{O}_3$
• D. $45.0$g conc. $HN{O}_3$

Answer 1

The correct option is D $45.0$g conc. $HN{O}_3$

Let us consider the problem:-

Molarity $\frac{weightofHN{O}_3}{massofHN{O}_3\times volumeofsolution\left(L\right)}$ (1)

Given that :

molarity of solution $=2$

volume of solution $=250ml=\frac{250}{1000}=\frac{1}{4}L$

Molar mass of

Since, weight of $HN{O}_3=molarity\times molarmass\times Volume\left(L\right)$

$=2\times 63\times \frac{1}{4}g=31.5$ gm

There fore,

It is the weight of $100\%$. $HN{O}_3$

But given aid $70\%$

Putting all values in equation (1)

weight $=31.5\times \frac{100}{70}g=45g$