A.n(O)= 3.2g 16g/mol =0.2mol,则3.2gO3所含电子数为0.2mol×8=1.6NA,故A错误;B.标准状况下,氯仿为液体,无法计算物质的量,故B错误;C.n= 0.4g 40g/mol =0.01mol,C3H4的结构简式为HC≡C-CH=CH2,0.01molC3H4含有共用电子对的数目为0.1NA,故C正确;D.Na2O2中阴离子为O22-,0.25molNa2O2中含有的阴离子数为0.25NA,故D错误.故选C.