Question

Can someone help me with this? Concentrated $ HN{O_3} $ has a specific gravity of $ 1.42 $ . It contains $ 69\% w/w $ (weight by weight) of $ HN{O_3} $ . calculate the molarity and molality of the solution?

Answer 1

Hint: From specific gravity, the density will be calculated. Using the density and volume of solution as one litre, gives the mass of solution. From the given weight by weight of $ HN{O_3} $ mass of nitric acid can be determined. It helps in determining the moles of $ HN{O_3} $ . Thus, moles of $ HN{O_3} $ and volume of solution given molarity. Later the mass of water in kilograms and moles of $ HN{O_3} $ gives molality.

Complete answer:
Given specific gravity of concentrated $ HN{O_3} $ is $ 1.42 $
 $ S = \dfrac{{{\rho _{HN{O_3}}}}}{{{\rho _{{H_2}O}}at{4^0}C}} $
By substituting the given specific gravity and water density at $ {4^0}C $ which is $ 1gm{l^{ - 1}} $
The density of nitric acid will be $ 1.42gm{l^{ - 1}} $
The mass of solution will be $ 1.42 \times 1000 = 1420gm $
Given $ 69\% w/w $ (weight by weight) of $ HN{O_3} $ it means $ 100gm $ of solution consists of $ 69gmHN{O_3} $
Thus, the solution consists of $ 1420 \times \dfrac{{69}}{{100}} = 979.8gHN{O_3} $
Nitric acid has molar mass of $ 63.01gmo{l^{ - 1}} $
Thus, moles of $ HN{O_3} $ will be $ \dfrac{{979.8}}{{63.01}} = 15.55moles $
Molarity is the ratio of number of moles of $ HN{O_3} $ which is solute divided by volume of solution in litres
 $ M = \dfrac{{15.55}}{1} = 15.6M $
Molarity of nitric acid is $ 15.6M $
Mass of solution is $ 1420gm $
Mass of nitric acid is $ 979.8g $
Thus, mass of water will be $ 1420 - 979.8 = 440.2g $
When the mass of water is converted into kilograms from mass,
Molality is the ratio of the moles of concentrated nitric acid to mass of water in kilograms
 $ m = \dfrac{{15.55}}{{440.2 \times {{10}^{ - 3}}}} = 35.3m $
The Molarity of nitric acid is $ 35.3m $ .

Note:
While calculating the molarity, the volume of solution must be in litres, if the volume of solution is in millilitres it should be multiplied by $ 1000 $ . While calculating the molality, the mass of the solvent must be in kilograms, if it is in grams, it should be converted into kilograms.