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NCERT Class XII Chemistry
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Question : 4 of 53
Marks: +1, -0
Concentrated nitric acid used in the laboratory is 68% nitric acid by mass in aqueous solution. What should be the molarity of such a sample of the acid if the density of the solution is 1.504 g mL −1^{-1} ?
Solution:  
Let mass of solution = 100 g
Then mass of nitric acid = 68 g
Molar mass of HNO3=63\mathrm{HNO}_3 = 63g mol−1^{-1}
Number of moles of HNO3=6863=1.079\mathrm{HNO}_3 = \frac{68}{63} = 1.079mole
Density of solution = 1.504 g mL−1^{-1}
∴ Volume of solution =100 g1.504 gmL−1= \frac{100\,\mathrm{g}}{1.504\,\mathrm{gmL}^{-1}} = 66.5 mL = 0.0665 L
Molarity of the solution =Number of moles of the soluteVolume of solution in L= \frac{\text{Number of moles of the solute}}{\text{Volume of solution in L}} =1.0790.0665= \frac{1.079}{0.0665} = 16.23 M
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