Answer to Question #156493 in General Chemistry for Maddy

Question

Masse of Li₂CO₃ necessary to prepare solution of 250,0ml with a 0,375mol/l concentration for Li⁺

Answer 1

1) n / 0.250 = 0.375 / 1, n = 0.09375 - the amount of substance in 250 ml.

2) n = n / 2, n = 0.046875, because one molecule of the sault dissociates, creating 2 Li+ ions

3) n = m / M, m = nM = 0.046875 * 73.891 = 3.46 (grams).

SOLUTION: 3.46 grams.