How many grams of concentrated nitric acid solution should be used to prepare `250 mL` of `2.0 M HNO_(3)`? The concentrated acid is `70% HNO_(3)`:

Options

A. `90.0 g conc. HNO_(3)`
B. `70.0 g conc. HNO_(3)`
C. `54.0 g conc. HNO_`
D. `45.0 g conc. HNO_(3)`

Answer

4

Explanation

First calculate the mass of `HNO_(3)` needed to prepare `250 mL` of `2.0 M HNO_(3)`.
`M=n_(HNO_(3))/V_(mL)xx(1000 mL)/L`
`=(m_(HNO_(3))//mm_(HNO_(3)))/V_(mL)xx(1000 mL)/L`
`2.0 mol L^(-1)=(m_(HNO_(3))//63g mol^(-1))/(250 mL)xx(1000 mL)/L`
Thus
`m_(HNO_(3))=3.15 g`
Since, the concentrated acid is `70% HNO_(3)`, the mass of concentrated nitric acid solution required to get `31.5 g HNO_(3)` will be
`(100 g)/(70 g)xx31.5 g`
`=45.0 g`