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2.4    Concentrated nitric acid used in laboratory work 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}\; ?

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According to given question, in 100 g of solution 68 g is nitric acid and rest is water.

So moles of 68 g HNO3 :-

                                                        =\ \frac{68}{63} = 1.08

Density of solution is given to be 1.504.

So volume of 100 g solution becomes :-  

                                                                       =\ \frac{100}{1.504} = 66.49\ mL

Thus, molarity of nitric acid is : 

                                                     Molarity = \frac{1.08}{\frac{66.49}{1000}} = 16.24\ M                                       

Posted by

Devendra Khairwa

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