Question

The total no. of ions present in 1mL of .1M barium nitrate Ba(NO3)2 solution is

Answer 1

Answer:

let no. of moles of barium nitrate be x

M= no. of moles of solute/ volume of solution in litre

1=moles of Ba(NO3)2×(1000)/1

1= 1000x

x = 0.001

so, if we dissociate barium nitrate we get two kind of ions i.e, barium ion and nitrate ion resp.

so to get total no. of ions in barium nitrate we should multiply moles of barium nitrate with 2

no. of ions=0.001×2=0.002