问题
选择题
在无土栽培中需用浓度为0.5mol/LNH4Cl、0.16mol/LKCl、0.24mol/LK2SO4的培养液，若用KCl、NH4Cl和（NH4）2SO4三种物质来配制1.00L上述营养液，所需三种盐的物质的量正确的是
A．0.4mol、0.5mol、0.12mol
B．0.66mol、0.5mol、0.24mol
C．0.64mol、0.5mol、0.24mol
D．0.64mol、0.02mol、0.24mol
答案
答案：D
0.5mol/LNH4Cl、0.16mol/LKCl、0.24mol/LK2SO4的混合溶液，设溶液体积为1L，所含离子的物质的量为NH4＋：0.5mol,K＋：0.64mol，Cl－：0.56mol,SO42－：0.24mol，若用KCl、NH4Cl和（NH4）2SO4三种物质来配制1.00L上述营养液,需用0.64mol KCl、0.02molNH4Cl、0.24mol（NH4）2SO4，故选D
