Question

在无土裁培中，配制500ml内含1mol /L NH₄Cl、0.32mol /L KCl、0.48mol/L K₂SO₄的某营养液，若用KCl、NH₄Cl和(NH₄)₂SO₄三种固体配制，则需此三种固体的物质的量分别为

A.0.40mol、0.50mol、0.12mol B.0.66mol、0.50mol、0.24mol

C.0.64mol、0.02mol、0.24mol D.0.64mol、0.50mol、0.24mol

Answer

C

【解析】

500ml内含1mol /L NH4Cl、0.32mol /L KCl、0.48mol/L K2SO4的营养液中，NH4Cl、KCl、K2SO4的物质的量分别为0.5mol、0.16mol、0.24mol，混合物中0.16mol+2×0.24mol=0.64mol；0.5mol+0.16mol=0.66mol；0.5mol，0.24mol。设KCl、NH4Cl和(NH4)2SO4三种固体分别为x、y、zmol；x=0.64，x+y=0.66，y+2z=0.5，z=0.24；得到KCl、NH4Cl和(NH4)2SO4三种固体分别为0.64mol，0.02mol；0.24mol。故选C。