用18.4mol?L-1的浓H2SO4配制100mL浓度为1mol?L-1的稀H2SO4,其操作可分为以下各步:A.用量筒取
\u5728100ml\u67d0\u6df7\u5408\u6eb6\u6db2\u4e2d,c(HNO3)=0.4mol/L,c(H2SO4)=0.1mol\u9700\u8981\u8003\u8651\u8fc7\u91cf\u3001\u4e0d\u8db3\u7684\u95ee\u9898
n(H+) = 0.4*0.1+0.1*0.1*2 = 0.06 mol
n(NO3-) = 0.04 mol
n(Cu) = 1.92/64 = 0.03 mol
\u6839\u636e\u65b9\u7a0b\u5f0f 3 Cu + 2NO3- + 8H+ = 3Cu2+ + 2NO + 4H2O
\u53ef\u89c1\uff0cH+\u4e0d\u8db3\uff0c\u8981\u6839\u636eH+\u8ba1\u7b97
n(Cu2+) = 0.06*3/8 = 0.0225 mol
\u6240\u4ee5 c(Cu2+) = 0.225 mol/L
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(1)溶液稀释前后物质的量不变,设需要浓硫酸的体积V,则18.4mol?L-1×V=1mol?L-1×100mL,V=5.4ml,定容时,将蒸馏水直接加入容量瓶,至液面接近环形刻度线1~2cm处,改用胶头滴管向容量瓶里逐滴滴入蒸馏水,至液面最低点恰好和刻度线相切,故答案为:5.4;胶头滴管;
(2)操作步骤有检查容量瓶是否漏水、量取、稀释、移液、洗涤、定容、摇匀等操作,倒数第二步为定容;
故答案为:E;
(3)量取的浓硫酸的体积是5.4mL,选用量筒的规格应稍大于量取液体的体积,所以选10mL的量筒;
故答案为:①;
(4)移液前应待溶液冷却到室温然后再移液;
故答案为:冷却;
(5)依据C=
n |
V |
A.容量瓶中原有少量蒸馏水,对n和V不产生影响,溶液的浓度不变,故A错误;
B.稀释所用烧杯洗涤2~3次,洗涤液一并注入容量瓶,正确操作,溶液浓度不变,故B错误;
C.定容时,仰视读刻度线,导致溶液的体积V偏大,溶液的浓度偏低,故C正确;
D.量取浓H2SO4时,仰视读数,量取的浓硫酸偏大,溶质的物质的量n偏大,溶液的浓度偏大,故D错误;
故选:C.
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