дано
m(ppa NaCL) = 200 g
W(NaCL) = 24%
+m(H2O) = 20 g
W1 -?
m(NaCL) = 200*24% / 100% = 48 g
m1(ppa NaCL) = 200+20 = 220 g
W1 = m(NaCL) / m1(ppa NaCL) * 100% = 48/220 * 100% = 21.82%
ответ 21.82%
дано
m(ppa NaCL) = 200 g
W(NaCL) = 24%
+m(H2O) = 20 g
W1 -?
m(NaCL) = 200*24% / 100% = 48 g
m1(ppa NaCL) = 200+20 = 220 g
W1 = m(NaCL) / m1(ppa NaCL) * 100% = 48/220 * 100% = 21.82%
ответ 21.82%