给出答案,请简要写出过程,若被采纳我一定会追加悬赏。某人须配制1L含0.5molNH4Cl、0.1molKCl、0.24molK2SO4溶液,若用KCl、lNH4Cl、(NH4)2SO4三种盐配制,则需三种盐的物质的量依次为:A.0.32mol、0.5mol、0.12molB.0.02mol、0.64mol、0.24molC.0.64mol、0.02mol、0.24molD.0.16mol、0.5mol、0.36mol真不好意思~是0.16MOLKCl!!!对不起了~