Gustina 35%-nog rastvora fosforne kiseline je 1,352 g/cm3. Kolika je koncentracija takvog rastvora?
w(H3PO4)=35%=0.35
ro=1.352 gcm-3
______________
c(H3PO4)=?
c=n/V=m/M*V=w(H3PO4)*m(otopine)/M(H3PO4)*V(otopine)=>kvocijent mase i volumena je ro(gustina), pa dobijemo :
c=w*ro/M(H3PO4)=0.35*1.352gcm-3/98gmol-1=4.82*10-3molcm-3=
4.82*moldm-3
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w(H3PO4)=35%=0.35ro=1.352
w(H3PO4)=35%=0.35
ro=1.352 gcm-3
______________
c(H3PO4)=?
c=n/V=m/M*V=w(H3PO4)*m(otopine)/M(H3PO4)*V(otopine)=>kvocijent mase i volumena je ro(gustina), pa dobijemo :
c=w*ro/M(H3PO4)=0.35*1.352gcm-3/98gmol-1=4.82*10-3molcm-3=
4.82*moldm-3