Question

Number of chloride ions present in 200 ml of 0.4 M
CaCl, solution is
2.4 x 1021
6.02 x 1020
9.6 x 1022
4.8 x 1021​

Answer 1

Answer:

$molarity = \frac{no \: of \: moles}{volume \: in \: litre}$

$0.4 = \frac{no \: of \: moles}{0.2}$

no of moles = 0.08

1 mole of CaCl2 contains 2 chloride ion

so, 0.08 moles of CaCl2 will have = 2×0.08×6.022×10²³ = 9.6×10²² chloride ions