Question

Calculate the concentration of nitric acid in moles per litre In a sample which has a density 1.41 L^-1
and mass percent of nitric acid in it being 69%.
​

Answer 1

Answer:

Answer:

given: 1.41 g/mL HNO3 solution 

the solution is 69% HNO3 

1. calculate the mass of HNO3 in the solution 

= 1.41 g/ml x 69% 

= 0.9729 g/ml 

2. convert the answer in 1 to mole quantity by dividing it with the molar mass of HNO3 which is 63 g/mol 

= 0.9729 g/ml divided 63 g/mol 

= 0.01544 mol/ml 

3. Since molar quantity is expressed as mole per L, you need to multiply the answer in 2 with 1000 (1L = 1000ml) 

= 0.01544 mol/ml X 1000 

= 15.44 M 

Answer 2

$\fbox{\texttt{\green{Answer:}}}$

Concentration of nitric acid in moles/litre = $\bold{15.44\:M}$

$\fbox{\texttt{\pink{Given:}}}$

• Mass percent of nitric acid is 69%. This means that 100 g of nitric acid solution contains 69 g of nitric acid by mass.
• Density of the sample = $\bold{1.41\:gL^{-1}}$

$\fbox{\texttt{\orange{To\:find:}}}$

Concentration of $\bold{HNO_3}$ in moles/litre in the sample.

$\fbox{\texttt{\red{How\:to\:Find:}}}$

No. of moles in 69 g $\bold{HNO_3=\frac{Mass}{Molar\:mass}}$

$\bold{\implies\dfrac{69}{63}=1.095\:mol}$

Volume of 100 g $\bold{HNO_3}$ solution

$\bold{=\frac{Mass}{Density}=\frac{100}{1.41}}$

$\implies\bold{70.92\:mL}$

$\implies\bold{0.07092\:L}$

$\hrulefill$

∴ Concentration of $\bold{HNO_3}$ in moles/litre

$\implies\bold{\dfrac{n}{v}=\dfrac{1.095}{0.07092}}$

$\implies\bold{15.44\:M}$