在一个密闭容器中，中间有一可自由滑动的隔板，将容器分成两部分．当左边充入28gN2，右边充入8gCO和CO2的混合气体时，隔板处于如图所示位置（两侧温度相同）．则混合气体中CO和CO2的分子个数比为（ ）
A．1：1
B．1：3
C．2：1
D．3：1
28gN2的物质的量为
=1mol．28g 28g/mol
左右两侧气体温度、压强相同，相同条件下，体积之比等于物质的量之比，令8gCO和CO2的混合气体的物质的量为n，则：
1mol：n=
：4 5
，解得，n=0.25mol．1 5
所以CO和CO2的混合气体的平均摩尔质量为
=32g/mol．8g 0.25mol
利用十字交叉法计算CO和CO2的物质的量之比，
所以CO和CO2的物质的量之比为12：4=3：1，
分子数之比等于物质的量之比，所以CO和CO2的分子数之比为3：1．
故选：D．
| 阅读理解。 | ||||||||
Children’s clubs | ||||||||
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