2020版高考浙江选考化学一轮教师备用题库：专题一 第三单元 溶液

教师备用题库

1.某溶质为Al2(SO4)3和Na2SO4的混合溶液中,c(Al3+)∶c(Na+)=1∶2,若c(S )=2.5 mol·L-1,则c(Al3+)为( ) A.0.5 mol·L-1 B.1 mol·L-1 C.2 mol·L-1

D.3 mol·L-1

- -

答案 B 根据溶液呈电中性可得,c(Na+)+3c(Al3+)=2c(S ),设c(Al3+)=x mol·L-1,则c(Na+)=2x mol·L-1,故2x mol·L-1+x mol·L-1×3=2c(S )=2.5 mol·L-1×2,解得x=1,故c(Al3+)=1 mol·L-1。

2.下列溶液中氯离子浓度与50 mL 1 mol·L-1氯化铝溶液中氯离子浓度相等的是( ) A.150 mL 1 mol·L-1的氯化钠溶液 B.75 mL 2 mol·L-1的氯化铵溶液 C.150 mL 3 mol·L-1的氯化钾溶液 D.75 mL 1 mol·L-1的氯化亚铁溶液

答案 C 根据化学式可知,50 mL 1 mol·L-1的AlCl3溶液中Cl-浓度为3 mol·L-1。150 mL 1 mol·LNaCl溶液中,Cl浓度为1 mol·L×1=1 mol·L,故A错误;75 mL 2 mol·L-1氯化铵溶液中,Cl-浓度为2 mol·L-1×1=2 mol·L-1,故B错误;150 mL 3 mol·L-1 KCl溶液中,Cl-浓度为3 mol·L-1×1=3 mol·L-1,故C正确;75 mL 1 mol·L-1 FeCl2溶液中,Cl-浓度为1 mol·L-1×2=2 mol·L-1,故D错误。

3.某溶质为CuSO4、Fe2(SO4)3、H2SO4的混合溶液100 mL,已知溶液中阳离子的浓度相同,且S 的物质的量浓度为6 mol·L-1,则此溶液最多溶解铁粉的质量为( ) A.11.2 g B.16.8 g C.19.6 g D.22.4 g

答案 D n(S )=0.1 L×6 mol·L-1=0.6 mol,已知溶质为CuSO4、Fe2(SO4)3、H2SO4的溶液中阳离子的浓度相同,则阳离子的物质的量也相同,设Cu2+、Fe3+、H+三种离子物质的量均为n,根据电荷守恒可知:2n+3n+n=0.6 mol×2,由此解得n=0.2 mol,Cu2+、H+、Fe3+都能与Fe反应生成Fe2+,最后溶质的成分为FeSO4,则n(FeSO4)=0.6 mol,根据铁原子守恒可知,此溶液最多溶解铁粉的物质的量为0.6 mol-0.2 mol=0.4 mol,则此溶液最多溶解铁粉的质量为0.4 mol×56 g·mol-1=22.4 g,故选D。

- --1

--1

-1

-

4.V L Fe2(SO4)3溶液中含有a g S ,取此溶液0.5V L,用水稀释至2V L,稀释后溶液中Fe3+的物质的量浓度是( ) A. mol·L-1 B. mol·L-1 C. mol·L-1 D. mol·L-1 答案 D a g硫酸根离子的物质的量为:硫酸根离子的物质的量为 mol×

.

·

-

= mol,取出的0.5V L溶液中含有-

= mol,稀释过程中溶质的物质的量不变,则

稀释后溶液中铁离子的物质的量为 mol× = mol,稀释后溶液中铁离子的物质的量浓度为

= mol·L-1,故选D。

5.配制一定物质的量浓度的NaCl溶液时,下列操作不正确的是( )

答案 C 称量固体时,当接近称量质量时,应用左手拿药匙,右手轻轻振动左手手腕,图示操作正确,故A不选;为防止溶解和洗涤时所用蒸馏水的体积超过容量瓶容积,溶解时应用量筒控制所加蒸馏水的量,图示操作正确,故B不选;移液时,玻璃棒应插在刻度线以下,图示操作错误,故选C;定容时,当液面接近刻度线1~2 cm时,用胶头滴管滴加蒸馏水至刻度线,图示操作正确,故D不选。

6.将一定质量的铁粉和铝粉的均匀混合物平均分成两份。一份与足量的稀硫酸反应,收集的H2在标准状况下的体积为1.792 L;另一份与某浓度的硝酸反应,生成的气体与标准状况下1.12 L的O2一起通入水中,反应后无气体剩余。则原混合物中铁粉的质量为( )

A.2.24 g B.3.36 g C.4.48 g D.5.60 g

答案 C 根据得失电子守恒有2n(Fe)+3n(Al)=

. . · -

×2,与某浓度的硝酸反应,

产生的气体与标准状况下1.12 L的O2一起通入水中,反应后无气体剩余,氧气得到的电子等于Al和Fe失去的电子,因此有3n(Fe)+3n(Al)=

. . · -

×4,解得n(Fe)=0.04

mol,混合物均匀分成两份,则混合物中铁质量为2×0.04×56 g=4.48 g,C项正确。 7.将15.66 g镁铝合金加入800 mL稀硝酸中,恰好完全反应(假设反应中还原产物只有NO),向所得溶液中加入足量的3 mol·L-1 NaOH溶液,测得生成沉淀的质量与原合金的质量相等,则下列有关叙述不正确的是 ( ) A.原稀硝酸的浓度为2.6 mol·L-1 B.生成NO的体积为11.648 L(标准状况) C.反应过程中共消耗1.56 mol NaOH D.合金中Al的质量分数约为58.6%

答案 C 将一定质量的镁铝合金加入稀硝酸中,两者恰好完全反应(假设反应中还原产物只有NO),发生反应:3Mg+8HNO3(稀)

3Mg(NO3)2+2NO↑+4H2O、Al+4HNO3(稀)

Al(NO3)3+NO↑+2H2O,向反应后的溶液中加入足量的NaOH溶液,发生反应:Mg(NO3)2+2NaOH

Mg(OH)2↓+2NaNO3、Al(NO3)3+4NaOH

NaAlO2+3NaNO3+2H2O。已

知沉淀为氢氧化镁,生成沉淀的质量与原合金的质量相等,则氢氧化镁中含有的氢氧根的质量与铝的质量相等,则合金中铝的质量为15.66 g× =9.18 g,镁的质量为15.66 g-9.18 g=6.48 g。镁的物质的量为

. ·

. · -

=0.27 mol,铝的物质的量为

=0.34 mol,根据反应方程式可知,硝酸的物质的量= ×0.27 mol+4×0.34 -

. .

mol=2.08 mol,则原稀硝酸的浓度为

=2.6 mol·L-1,A项正确;根据上述分析,生成

NO的物质的量为 ×0.27 mol+0.34 mol=0.52 mol,体积为11.648 L(标准状况),B项正确;反应过程中消耗NaOH的物质的量为2×0.27 mol+4×0.34 mol=1.9 mol,C项错误;合金中Al的质量分数为 . ×100%≈58.6%,D项正确。

8.为研究某(NH4)2SO4和NH4HSO4混合物样品的组成,称取四份该样品,分别逐滴加入相同浓度的NaOH溶液25.0 mL,加热并完全反应,产生NH3的体积(NH3的体积已折算成标准状况,不考虑NH3在水中的溶解)如下表所示:

实验序号 Ⅰ Ⅱ Ⅲ Ⅳ

.

NaOH溶液体积(mL) 25.0 25.0 25.0 25.0 样品质量(g) NH3体积(mL) 10.83.62 5.43 7.24 6 896 1 344 1 344 896

(1)样品中(NH4)2SO4和NH4HSO4的物质的量之比为 。 (2)NaOH溶液中NaOH的物质的量浓度为 。 答案 (1)1∶2 (2)4.00 mol/L 解析 (1)NH4HSO4的电离方程式为NH4HSO4是发生反应:H++OH- + N +H+S ,加入NaOH溶液的过程中先

-

H2O,然后发生反应:N +OH-

NH3↑+H2O。按Ⅰ组数据计算,设

.

. .

n[(NH4)2SO4]和n(NH4HSO4)分别为x mol、y mol,依题意有

,解得

x=0.01、y=0.02,所以n[(NH4)2SO4]∶n(NH4HSO4)=0.01 mol∶0.02 mol=1∶2。 (2)应选NaOH已完全反应的第Ⅳ组,10.86 g样品中n(H+)=n(NH4HSO4)=0.02 mol×3=0.06 mol,据H++OH- H2O知,消耗0.06 mol NaOH,据反应N +OH- .

NH3↑+H2O知,生成

0.896 L NH3消耗NaOH的物质的量: . =0.04 mol,NaOH溶液中NaOH的物质的量浓度为

. .

.

=4.00 mol/L。

9.(2017浙江4月选考,29,4分)分别称取2.39 g(NH4)2SO4和NH4Cl固体混合物两份。

(1)将其中一份配成溶液,逐滴加入一定浓度的Ba(OH)2溶液,产生的沉淀质量与加入Ba(OH)2溶液体积的关系如图。混合物中n[(NH4)2SO4]∶n(NH4Cl)为 。

(2)另一份固体混合物中N 与Ba(OH)2溶液(浓度同上)恰好完全反应时,溶液中

c(Cl-)= (溶液体积变化忽略不计)。 答案 (1)1∶2 (2)0.100 mol·L-1 解

析

(1)n[(NH4)2SO4]=n(BaSO4)=

. · -

=0.01

mol,n(NH4Cl)=

. - . · -

. · -

=0.02 mol,n[(NH4)2SO4]∶n(NH4Cl)=1∶2。

(2)c[Ba(OH)2]×0.1 L=n[Ba(OH)2]=n(BaSO4)=0.01 mol,c[Ba(OH)2]=0.1 mol·L-1。

2N ~ Ba(OH)2

2 1

(0.01×2+0.02)mol 0.1 mol·L-1×V =( . . )

. · -

解得V=0.2 L,c(Cl-)=0.02 mol÷0.2 L=0.100 mol·L-1。

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