Calculate the molarity of 1.8 g HNO3 dissolved in 250 mL aqueous solution.

Mass of HNO₃ = 1.8 g,

Volume of solution = 250 mL = 0.250 L

To find : 

Molarity

Formulae :

i. Number of moles = \(\frac{Mass\,of\,substance\,in\,grams}{Molar\,mass\,of\,substance}\)

ii. Molarity (M) = \(\frac{Number\,of\,moles\,of\,solute}{Volume\,of\,solution\,in\,litres}\)

Calculation : 

Molar mass of HNO₃ = 63 g mol^-1

Using formula (i),

Number of moles = \(\frac{Mass\,of\,substance\,in\,grams}{Molar\,mass\,of\,substance}\) 

= \(\frac{1.8}{63}\)

= 0.0286 mol 

Using formula (ii),

Molarity (mol/L) = \(\frac{Number\,of\,moles\,of\,solute}{Volume\,of\,solution\,in\,litres}\)

= \(\frac{0.0286\,mol}{0.250L}\) 

= 0.114 M

∴ Molarity of the HNO₃ solution = 0.114 M