Question

How many grams of the concentrated nitric acid solution is required to prepare $250mL$ of $2.0M$ nitric acid solution? (Concentration of $H{NO}_3$ is $70\%$)

• A. $45g$
• B. $90g$
• C. $70g$
• D. $54g$

Answer 1

The correct option is A $45g$

The required number of moles= $=\left(\frac{250}{1000}\right)\times 2=0.5$moles

(no. of moles=molarity × volume)
so,required mass of $HN{O}_3=0.5\times 63=31.5grams$
(given mass=no. of moles × molar mass)

Given,

70 grams of $HN{O}_3$ are present in 100 grams of the solution,

so,1 gram will be present in $\frac{100}{70}$ grams of solution
hence 31.5 grams will be present in $\frac{100}{70}\times 31.5$ grams of solution.
so the amount of concentrated nitric acid solution used is 45 grams