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