1.

Calcualte the concentration of nitric acid in moles per litre in a sample which has a density, 1.41 g mL^(-1) and mass per cent of nitric in it being 69%.

Answer»

Solution :Mass PERCENT of `69%` means that 100 G of nitric acid solution contain 69 g of nitric acid by mass.
Molar mass of nitric acid `(HNO_(3))=1+14+48=63gmol^(-1)`
`therefore"Moles in 69 g "HNO_(3)=(69g)/(63gmol^(-1))=1.095" mole"`
`"Volume of 100 g nitirc acid solution"=(100G)/(1.41g mL^(-1))=70.92mL=0.07092L`
`therefore"CONC. of "HNO_(3)" in moles per litre"=("1.095 mole")/("0.07092 L")=15.44M`


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