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+)*NA = 0.015*6.02*1023 = 9.03*1021
correct answer is d
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