化学
标准状况下有①0.112L水;②3.01×1023个氯化氢分子;③13.6g H2S气体;④0.2mol氨气,下列对这四种物质的关系由小到大排列正确的是(  )A. 体积:①③②④B. 密度:④①③②C. 质量:①④③②D. 氢原子数:②④③①

2019-11-26

标准状况下有①0.112L水;②3.01×1023个氯化氢分子;③13.6g H2S气体;④0.2mol氨气,下列对这四种物质的关系由小到大排列正确的是(  )
A. 体积:①③②④
B. 密度:④①③②
C. 质量:①④③②
D. 氢原子数:②④③①
优质解答
标况下,水是液体,气体摩尔体积对其不适用,n(HCl)=
3.01×1023
6.02×1023/mol
=0.5mol、n(H2S)=
13.6g
34g/mol
=0.4mol、n(NH3)=0.2mol,
A.水的体积是0.112L,V(HCl)=0.5mol×22.4L/mol=11.2L、V(H2S)=0.4mol×22.4L/mol=8.96L、V(NH3)=0.2mol×22.4L/mol=4.48L,所以由小到大排列正确的是①④③②,故A错误;
B.水的密度是1g/mL,根据ρ=
M
Vm
知,气体密度与其摩尔质量成正比,HCl的摩尔质量是36.5g/mol、硫化氢摩尔质量是34g/mol、氨气摩尔质量是17g/mol,气体密度都小于1g/mL,所以密度从小到大顺序是④③②①,故B错误;
C.水的质量是112g,m(HCl)=0.5mol×36.5g/mol=18.25g、m(H2S)=0.4mol×34g/mol=13.6g、m(NH3)=0.2mol×17g/mol=3.4g,所以质量从小到大顺序是④③②①,故C错误;
D.水中N(H)=
112g
18g/mol
×NA/mol
×2=12.4NA,HCl中N(H)=0.5mol×NA/mol×1=0.5NA,硫化氢中N(H)=0.4mol×NA/mol×2=0.8NA,氨气中N(H)=0.2×NA/mol×3=0.6NA,所以H原子个数由小到大顺序是②④③①,故D正确.
故选D.
标况下,水是液体,气体摩尔体积对其不适用,n(HCl)=
3.01×1023
6.02×1023/mol
=0.5mol、n(H2S)=
13.6g
34g/mol
=0.4mol、n(NH3)=0.2mol,
A.水的体积是0.112L,V(HCl)=0.5mol×22.4L/mol=11.2L、V(H2S)=0.4mol×22.4L/mol=8.96L、V(NH3)=0.2mol×22.4L/mol=4.48L,所以由小到大排列正确的是①④③②,故A错误;
B.水的密度是1g/mL,根据ρ=
M
Vm
知,气体密度与其摩尔质量成正比,HCl的摩尔质量是36.5g/mol、硫化氢摩尔质量是34g/mol、氨气摩尔质量是17g/mol,气体密度都小于1g/mL,所以密度从小到大顺序是④③②①,故B错误;
C.水的质量是112g,m(HCl)=0.5mol×36.5g/mol=18.25g、m(H2S)=0.4mol×34g/mol=13.6g、m(NH3)=0.2mol×17g/mol=3.4g,所以质量从小到大顺序是④③②①,故C错误;
D.水中N(H)=
112g
18g/mol
×NA/mol
×2=12.4NA,HCl中N(H)=0.5mol×NA/mol×1=0.5NA,硫化氢中N(H)=0.4mol×NA/mol×2=0.8NA,氨气中N(H)=0.2×NA/mol×3=0.6NA,所以H原子个数由小到大顺序是②④③①,故D正确.
故选D.
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