已知氢元素有1H、2H、3H三种同位素,氯元素有35Cl、37Cl两种同位素.

已知氢元素有1H2H3H三种同位素,氯元素有35Cl37Cl两种同位素.由这五种微粒构成的HCl分子中,其相对分子质量可能有()

    A             1                B 5             C 6  D 7

答案

考点  相对分子质量及其计算.

专题  计算题.

分析:  采用排列组合的方法计算存在的分子种类=C31×C21,然后去掉相对分子质量相同的即可.

解答:  解:H的核素有3种,氯的核素有2种,所以HCl的种类=C31×C21=6种,但11H1737Cl13H1735Cl的相对分子质量相同,所以HCl分子的相对分子质量数值可能有5种,

故选B

点评:  本题以分子的构成为载体考查了同位素,难度不大,根据排列组合分析解答即可.

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