Question

How many lithium ions are present in 25.0 mL of 0.300 M Li2CO3 solution?

a. 9.03 x 1022 ions

b. 4.52 x 1021 ions

c. 0.0150 ions

d. 9.03 x 1021 ions

Answer 1

C(Li₂CO₃) = 0.3 M

V(solution) = 25 mL = 0.025 L

n(Li₂CO₃) = C*V = 0.0075 mol

n(Li⁺) = 2n(Li₂CO₃) = 0.0075*2 = 0.015 mol (because Li₂CO₃ has 2 lithium ions in its formula)

N(Li⁺) = n(Li⁺)*N_A = 0.015*6.02*10²³ = 9.03*10²¹

correct answer is d