Answer: 2.9
Explanation:
Molar mass of Cu(NO3)2 = 187.5 g/mol
Molar mass of water = 18.0 g/mol
%H2O = (5.0 g - 3.9 g) / 5.0 g X 100 (100 is to get percent) = (1.1 g) / 5.0 g X 100
= 22.0 %
% H2O = ( mass of H20 alone) / (mass of total hydrate)
0.22 = 18.0n / (18.0n + 187.5)
n = 2.9 = (about) 3
When a 5.0g sample of Cu(NO₃)₂ . nH₂O is heated, and 3.9g of the anhydrous salt remains. The value of n is 3.
Water mass = Hydrate mass - Anhydrous mass
= 5.0 - 3.9
= 1.1 g water
Number of moles = [tex]\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Now find the moles of H₂O and moles of Cu(NO₃)₂.
Moles of H₂O = [tex]\frac{\text{Mass of water}}{\text{Molar mass of water}}[/tex]
= [tex]\frac{1.1\ g}{18\ g/mol}[/tex]
= 0.061 moles
Moles of Cu(NO₃)₂ = [tex]\frac{\text{Mass of}\ Cu(NO_{3})_{2}}{\text{Molar mass of}\ Cu(NO_{3})_{2}}[/tex]
= [tex]\frac{3.9\ g}{187.5\ g/mol}[/tex]
= 0.0208 moles
Ratio of moles of H₂O and moles of Cu(NO₃)₂.
= [tex]\frac{0.061}{0.0208}[/tex]
= 2.9
≈ 3
The value of n is 3. So the formula will be Cu(NO₃)₂ . 3H₂O.
Thus, from above conclusion we can say that when a 5.0g sample of Cu(NO₃)₂ . nH₂O is heated, and 3.9 g of the anhydrous salt remains. The value of n is 3.
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