Question

将标准状况下的11.2L氨气溶于100L水中，得到密度为0.868g．cm负三次方的氨水。求所得氨水中溶质的物质...

将标准状况下的11.2L氨气溶于100L水中，得到密度为0.868g．cm负三次方的氨水。求所得氨水中溶质的物质的量浓度。

Answer 1

n(NH3) = 11.2/22.4 = 0.5mol
m(NH3) = 17*0.5 = 8.5g
m(溶液) = 8.5 + 100*10^3 = 100008.5g
V(溶液) = 100008.5/0.868 = 115217ml = 115.2L
c(NH3) = 0.5/115.2 = 4.34*10^(-3) mol/L

Answer 2

是100毫升水吧。
n(NH3) = 11.2/22.4 = 0.5mol
m(NH3) = 17*0.5 = 8.5g
m(溶液) = 8.5 + 100 = 108.5g
V(溶液) = 108.5/0.868 =125mL=0.125L
c(NH3) = 0.5/0.125=4mol/L