Вопрос школьника по предмету Химия
К 450 мл 5 M раствора H2SO4 прилили 150 мл воды , получили раствор с плотностью 1,220г, ил вычислить массовую долю серной кислоты и эквивалентную концентрацию
Ответ учителя по предмету Химия
Дано:
Cm = 5 M
V1(p-pa) = 450 мл
+m(H2O) = 150 мл
p2 = 1,22г/мл
Найти:
w(H2SO4), Cn
Решение:
w = m(в-ва) ÷ m(p-pa)
m(H2SO4) = n(H2SO4) × Mr(H2SO4)
n(H2SO4) = V1(p-pa) × Cm = 0,45л × 5 моль/л = 2,25 моль
Mr(H2SO4) = 98 г/моль
m(H2SO4) = 2,25 моль × 98 г/моль = 220,5 г
m(p-pa) = p2 × V2
V2 = V1 + V(H2O) = 450мл + 150мл = 600мл
m(p-pa) = 1,22 г/мл × 600 мл =
732 г
w(H2SO4) = 220,5 г ÷ 732 г = 0,30 × 100% = 30%
Сn = Cm × z
z(H2SO4) = 2
Cn = 5 M × 2 = 10 N
Ответ: w(H2SO4) = 30%, Cn = 10 N
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