Дано
w1(MgCl2) = 3% = 0.03
w2(MgCl2) = 4.59% = 0.0459
m(MgCl2) = 2 г
Решение
w2(MgCl2) = (w1(MgCl2)m1(MgCl2) + m(MgCl2)) / (m1(MgCl2) + m(MgCl2)) = (0.03m1(MgCl2) + 2) / (m1(MgCl2) + 2) = 0.0459
(0.03m1(MgCl2) + 2) / (m1(MgCl2) + 2) = 0.0459
0.03m1(MgCl2) + 2 = 0.0459(m1(MgCl2) + 2)
0.03m1(MgCl2) + 2 = 0.0459m1(MgCl2) + 0.0918
0.03m1(MgCl2) + 1.9082 = 0.0459
0.0159m1(MgCl2) = 1.9082
m1(MgCl2) = 120 г
Ответ: 120 г