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Determinati cate grame de acid oxalic se gasesc in 250ml solutie de concentratie 0,25M (AH=1, AC=12,AO=16)

Răspuns :

HOOC-COOH=C2H2O4 ==>MC2H2O4=2AC+2AH+4AO=12*2+2*1+4*16=90g/mol
Vs=250mL=0.25L
Cm=0.25 moli/L

Cm=md/M*Vs==>md=Cm*M*Vs=0.25*0.25*90=5.625g C2H2O4

sau 

Cm=n/Vs==>n=Cm*Vs=0.25*0.25=0.0625 moli C2H2O4

n=m/M==>m=n*M=0.0625*90=5.625g C2H2O4