Question

27.
The total number of ions persent in 1 mL of 0.1 M
barium nitrate Ba(NO3), solution is -
(1) 6.02 x 1018
(2) 6.02 x 1019
(3) 3.0 x6.02 x1019 (4) 3.0 x6.02 x1018
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Answer 1

Molarity of Ba(NO3)2 solution = 0.1 mol/L

Volume of solution = 1ml = 0.001L

Moles of Ba(NO3)2 = M x V = 0.1 x 0.001 = 1 x 10^-4 mol

I mole of Ba(NO3)2 = 6.02 x 10^23 molecules

Total no. of ions present in 1 mole of Ba(NO3)2 = 3 x 6.02 x 10^23

Total no. of ions present in 1 mL of 0.1 M  Ba(NO3) solution = 3 x 6.02 x 10^23 x 10^-4 = 3 x 6.02 x 10^19 ions

Option (3) is correct.