Sa koliko molekula kristalne vode kristalise kupri-nitrat, ako se zarenjem iz o.996g kristalohidrata dobija 0.77g bezvodne soli?
m(KH) = 0.996 gm(BV) = 0.770 g_______________
x (H2O) = ?
Cu(NO3)2.XH2O --> Cu(NO3)2 + xH2O
Cu(NO3)2.XH2O - Kristalohidrat (KH)Cu(NO3)2 - Bezvodna so (BV)
n(KH) : n(BS) = 1 : 1 --> n(KH) = n(BV)
m(KH)/M(KH) = m(BV)/M(BS)
M(KH) = M(BV)*m(KH)/m(BV), M(BV)=187.456 g/mol
= 187.456*0.996/0.77 = 242.59 g/mol
M(KH) = M(BS) + xM(H2O)
X = (M(KH)-M(BS))/M(H2O) = (242.59 - 187.456)/18
X = 3
Cu(NO3)2.3H2O
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Broj molekula
m(KH) = 0.996 g
m(BV) = 0.770 g
_______________
x (H2O) = ?
Cu(NO3)2.XH2O --> Cu(NO3)2 + xH2O
Cu(NO3)2.XH2O - Kristalohidrat (KH)
Cu(NO3)2 - Bezvodna so (BV)
n(KH) : n(BS) = 1 : 1 --> n(KH) = n(BV)
m(KH)/M(KH) = m(BV)/M(BS)
M(KH) = M(BV)*m(KH)/m(BV), M(BV)=187.456 g/mol
= 187.456*0.996/0.77 = 242.59 g/mol
M(KH) = M(BS) + xM(H2O)
X = (M(KH)-M(BS))/M(H2O) = (242.59 - 187.456)/18
X = 3
Cu(NO3)2.3H2O