these piglets are cute.改感叹句

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• VmLAl2(SO4)3溶液中含Al3+ag,取V/4mL溶液稀释到4VmLVmLAl2(SO4)3溶液中含Al3+ag,取V/4mL,并加水稀释至4VmL,则稀释后SO42-物质的量浓度为（ ）答案是125a/36V mol/L可是我是这么算的由已知可得Al2(SO4)3的质量为a/2 g.所以它的物质的量为a/2/342=a/684 mol.所以这个溶液的物质的量浓度就是a/684/V=a/684V.根据v1c1=v2c2可得 a/684V·V/4=C·4V 解出来C不是正确答案那个数.可是我觉得我这么算也对啊..怎么回事不对，解出C来是稀释之后溶液的量浓度，然后再用那个量浓度乘3，就是题目中要求的了。可是算出那个数也不是正确答案。