Calculate the concentration of nitric acid in moles per litre in a sample which has a density.1.41 g/ml and the mass Y. of nitric acids is 69Y.

Open in App

Solution

Given $-$ Mass of ${H N O}_{3}$ in solution $= 69 g$
Hence, mass of $H_{2} O = 31 g$
Numbers of moles of ${H N O}_{3} = 69 / 63 = 1.09 m o l e s$
Volume $= \frac{M a s s}{D e n s i t y} = \frac{100}{1.41} = 70.9 m l$
Concentration $(M) = \frac{N o . o f m o l e s o f s o l u t e}{V o l u m e o f s o l u t i o n}$
$= \frac{1.09}{70.9} \times 100 = 1.5 M$

Suggest Corrections

Similar questions

Q. Calculate the concentration of nitric acid in moles per litre in a sample which has a density.

1.41

m L^{- 1}

and the mass per cent of nitric acid in it being

69 %

Calculate the concentration of nitric acid in moles per litre in a sample which has a density, 1.5 g per mL and the mass percent of nitric acid in it being 69%.

Join BYJU'S Learning Program

Calculate the concentration of nitric acid in moles per litre in a sample which has a density.1.41 g/ml and the mass Y. of nitric acids is 69Y.

Given − Mass of HNO3 in solution =69g Hence, mass of H2O=31gNumbers of moles of HNO3=69/63=1.09molesVolume =MassDensity=1001.41=70.9mlConcentration (M)=No.ofmolesofsoluteVolumeofsolution =1.0970.9×100=1.5M