B
分析:A.鋁與足量氫氧化鈉反應(yīng),1個鋁原子失去3個電子變成+3價的偏鋁酸根離子;
B.乙酸的化學(xué)式為CH
3COOH,屬于羧酸,官能團為羧基-COOH;
C.標(biāo)準(zhǔn)狀況下苯為液態(tài),不能用Vm=22.4L/mol;
D.膽礬的摩爾質(zhì)量為250g/mol.
解答:A.5.4g鋁的物質(zhì)的量為n=
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=
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=0.2mol,2Al+2NaOH+2H
2O=2NaAlO
2+3H
2↑由方程式可以看出,反應(yīng)關(guān)系式為2Al~3H
2~6e
-,所以5.4g鋁與足量氫氧化鈉反應(yīng)轉(zhuǎn)移的電子數(shù)為0.6N
A,故A錯誤;
B.根據(jù)乙酸的結(jié)構(gòu)簡式CH
3COOH,知:乙酸官能團為羧基-COOH,1 mol 分子中含有碳?xì)鋯捂I為4 mol,含碳碳單鍵為1 mol,含碳氧雙鍵為1 mol,1 mol碳氧雙鍵含2mol共價鍵,含碳氧單鍵為1 mol,所以1 mol乙酸分子中共價鍵總數(shù)為8N
A,故B正確;
C.標(biāo)準(zhǔn)狀況下苯為液態(tài),不能用n=
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計算,所以標(biāo)準(zhǔn)狀況下,11.2 L苯所含原子數(shù)不為6N
A,故C錯誤;
D.160g膽礬的物質(zhì)的量為n=
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=
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=
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mol,膽礬溶于水溶質(zhì)為硫酸銅,所以硫酸銅溶液的物質(zhì)的量濃度為C
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=
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≠1 mol/L,故D錯誤;
故選B.
點評:本題考查了阿伏加德羅常數(shù)的有關(guān)計算,掌握相關(guān)的公式并能熟練運用是解答的關(guān)鍵,題目較簡單.