Question

calculate concentration of nitric acid in moles per litre in a sample having density 1.41gmL^-1 ,mass percent of nitric acid being 69%​

Answer 1

Solution :

Mass percent of nitric acid in the sample = 69 % [Given]

Thus, 100 gof nitric acid contains 69 g of nitric acid by mass.

Molar mass of nitric acid (HNO_3)

= (1 + 14 + 3(16)) g mol^-1

= 1 + 14 + 48

= 63 g mol^-1

Number of moles in 69 g of HNO,

= 69g/63g mol^-1

= 1.095 mol

Volume of 100g of nitric acid solution

= Mass of solution/density of solution

= 100g/1.41gmL

= 70.92 mL

= 70.92 x10^-3 L

Concentration of nitric acid

= 1.095 mole/70.92x10^-3L

= 15.44 mol/L

=> Concentration of nitric acid = 15.44 mol/L

Answer 2

Question:-

Calculate the concentration of nitric acid in moles per litre in a sample having density 1.41gm / ltr ,mass percent of nitric acid being 69%. In other words, find Molarity (M) ?

Answer:-

Concentration in moles per litre = Molarity (M) = (No. of moles of solute) / (Volume of solution in litres)

And No. of moles of solute = (Mass of solute) / (Molecular Mass of solute)

So,

Molarity = (Mass of solute) / [ Molecular mass of solute * Volume of solution of litres ]

Given: 100% of the sample has 69% HNO₃ by mass

=> 100g of the sample has 69g of HNO₃

So, Mass of solute = 69g

and Mass of Solution = 100g

Molecular mass of HNO₃ :-

= 1 + 14 + (16 * 3) g/mol

= 63 g/mol

So, Molecular mass of solute = 63 g/mol

Volume of solution :-

(Mass of solution) / (Density of Solution)

= (100g) / (1.41 g/ml)

= 70.92 ml

We have to convert ml to litres

= 70.92 * (10)^(-3) l

So,

Molarity = (Mass of solute) / [ Molecular mass of solute * Volume of solution of litres ]

= (69) / (63 * 70.92 * (10)^(-3) M

= 15.44 M

(M stabds for molar)

Ans. Molarity = 15.44 M