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 .

Correct answer: A. 45.0 ~g conc. HNO_3 Given, molarity of solution =2 Volume of solution =250 ~mL=(250)/(1000)=(1)/(4) ~L Molar mass of HNO_ 3 =1+14+3 ×16=63 ~g ~mol^ -1 Molarity aligned & = weight of HNO_3 mass of .HNO_3 × volume of solution ( L) & Weight of HNO_ 3 = molarity × mol. mass & × volume (L) & =2 ×63 × (1)/(4) ~g=31.5 ~g & aligned It is the weight of100 \% HNO_3 . But the given acid is 70 \% HNO_3. Its weight =31.5 × (100)/(70)~g=45 ~g

NEET 2013

Exam: NEET
Year: 2013
Subject: Chemistry
Chapter: Solutions

NEET 2013 · Chemistry · Solutions

How many grams of concentrated nitric acid solution should be used...

How many grams of concentrated nitric acid solution should be used to prepare

250 \mathrm{~mL}

2.0 \mathrm{M} \mathrm{HNO}_3

? The concentrated acid is

70 \% \mathrm{HNO}_3

Options

A $45.0 \mathrm{~g}$ conc. $\mathrm{HNO}_3$
B $90.0 \mathrm{~g}$ conc. $\mathrm{HNO}_3$
C $70.0 \mathrm{~g}$ conc. $\mathrm{HNO}_3$
D $54.0 \mathrm{~g}$ conc. $\mathrm{HNO}_3$

Correct answer

A. $45.0 \mathrm{~g}$ conc. $\mathrm{HNO}_3$

Step-by-step solution

Given, molarity of solution

=2

Volume of solution

=250 \mathrm{~mL}=\frac{250}{1000}=\frac{1}{4} \mathrm{~L}

Molar mass of

\mathrm{HNO}_{3}=1+14+3 \times16=63 \mathrm{~g} \mathrm{~mol}^{-1}\because Molarity

\begin{aligned} & =\frac{\text{ weight of } \mathrm{HNO}_3}{\text{ mass of } \left.\mathrm{HNO}_3 \times \text{ volume of solution ( } L\right)}\\ & \therefore \text{ Weight of } \mathrm{HNO}_{3}=\text{ molarity } \times \text{ mol. mass }\\ & \times \text{ volume }(\mathrm{L})\\ & =2 \times63 \times \frac{1}{4} \mathrm{~g}=31.5 \mathrm{~g}\\ & \end{aligned}

It is

the weight of100 \%

\mathrm{HNO}_3

. But the given acid is

70 \% \mathrm{HNO}_3.

\therefore

\text{ Its weight }=31.5

\times

\frac{100}{70}\mathrm{~g}=45 \mathrm{~g}

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