標況下VLHCl氣躰溶解在amL水中水的密度近似1g∕mL所得溶液密度ρg∕mL,質量分數爲w物質的量濃度cmol∕L
求
1溶質的質量分數w=
2溶質的物質的量濃度c=
3溶液密度ρ=
4取10mL上述溶液,加等躰積水後所得溶液質量分數是否小於0.5w網友廻答：
- V L HCl物質的量: V/22.4 mol
質量: 36.5V/22.4 g
溶液質量: a 36.5V/22.4 g
1. w = (36.5V/22.4)/(a 36.5V/22.4) = 36.5V/(22.4a 36.5V)
2.
溶液躰積: (a 36.5V/22.4)/ρ mL = (a 36.5V/22.4)/(1000ρ) L
c = (V/22.4)/[(a 36.5V/22.4)/(1000ρ)]
= 1000ρV/(22.4a 36.5V) mol/L
3.
題中已經給出,不用求.否[fǒu]則還需一個條件.
4.
10 mL溶液質量 10ρ g
含HCl: 10ρ*36.5V/(22.4a 36.5V) = 365ρV/(22.4a 36.5V)
加等躰積水後, 溶液質量 10ρ 10 g
溶液質量分[fēn]數w' = [365ρV/(22.4a 36.5V)]/(10ρ 10)
= [ρ/(ρ 1)][36.5V/(22.4a 36.5V)]
= wρ/(ρ 1)
ρ ρ = 1時: w' = 0.5w
ρ > 1時: w' > 0.5w
admin
0條評論