标准状况下将VLHCl溶解在1L水中(水的密度近似为1gmL),所得溶液的密度为ρgmL,质量分数为w,物质浓度

2025-04-30 13:07:33
推荐回答(1个)
回答1:

A.VL HCl的物质的量=

VL
22.4L/mol
=
V
22.4
mol,其质量=
V
22.4
mol×36.5g/mol=
36.5V
22.4
g,1L水的质量=1000mL×1g/mol=1000g,故溶液质量=(
36.5V
22.4
+1000)g,溶液体积=
V
22.4
mol
c mol/L
=
V
22.4c
L,故溶液密度ρ=
(
36.5V
22.4
+1000)g
V
22.4c
×103mL
=
(36.5V+22400)c
1000V
g/mL,故A错误;
B.VL HCl的物质的量=
VL
22.4L/mol
=
V
22.4
mol,其质量=
V
22.4
mol×36.5g/mol=
36.5V
22.4
g,1L水的质量=1000mL×1g/mol=1000g,故溶液质量=(
36.5V
22.4
+1000)g,溶液体积=
(
36.5V
22.4
+1000)g
1000ρ g/L
=
36.5V+22400
22400ρ
L,故溶液浓度c=
V
22.4
mol
36.5V+22400
22400ρ
L
=
1000ρV
36.5V+22400
mol/L,故B正确;
C.VL HCl的物质的量=
VL
22.4L/mol
=
V
22.4
mol,其质量=
V
22.4
mol×36.5g/mol=
36.5V
22.4
g,1L水的质量=1000mL×1g/mol=1000g,故溶液质量=(
36.5V
22.4
+1000)g,则:(
36.5V
22.4
+1000)g×w=
36.5V
22.4
g,解得V=
22400w
36.5(1?w)
,故C正确;
D.根据c=
1000ρw
M
可知,ρ=
36.5c
1000w
,故D正确,
故选A.