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. 45.0g conc . `HNO_(3)`
B. 90.0g conc.` HNO_(3)`
C. 70 .0 g conc .` HNO_(3)`
D. 54.0g conc . `HNO_(3)`

Answer

A

Explanation

Given , molarity of solution =2
Volume of solution =250 mL `=(250)/(1000)=(1)/(4)L`
Molar mass of
`HNO_(3)=1+14+3xx16=63g "mol" ^(-1)`
`:'` Molarity
`=("Weight of " HNO_(3))/("Molecular mass of " HNO_(3)xx "volume of solution (L)")`
`:.` Weight of `HNO_(3)` = molarity `xx` molecular mass `xx` volume (L)
`=2xx63xx(1)/(4)g=31.5 g`
It is weight of `100%HNO_(3)`
But the given acid is `70%HNO_(3)`
`:.` Its weight `=31.5xx(100)/(70)g=45g`