ATAR Notes: Forum
VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Chemistry => Topic started by: chemmy on March 13, 2012, 04:12:00 pm
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If someone could show me how to do the following questions then it would be very very very much appreciated! :)
1. A mixture of ZnSO4 and ZnSO4.6H20 has a mass of 36.5 g. Upon heating the water is driven off and the mass of the ZnSO4 remaining is 28.1 g. Determine the % by weight of the ZnSO4.6H20 in the original mixture.
2. A mixture of Cu(NO3)2 and Cu(NO3)2.2.5H20 has a mass of 3.150 g. Upon heating the water is driven off and the mass of the Cu(NO3)2 is 2.853 g. Determine the % by weight of the Cu(NO3)2.2.5H20 in the original mixture
3. The mass (g) of one 19F2 molecule is closest to:
(Use the isotopic mass of 19F = 18.9984 amu.)
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1. m(H2O) = 36.5 - 28.1 = 8.4g
n(H2O) = 8.4/18 = 0.467mol
n(H2O) = n(ZnSO4.6H2O) x 1/6 = 0.0778mol
m(ZnSO4.6H2O) = 0.0778 x M = 20.96g
%w/w = 20.96/36.5 x 100 = 57.4%w/w
2. I don't know what Cu(NO3)2.2.5H2O is... no subscripts
3. mass = 38/6.02x10^23 g
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If someone could show me how to do the following questions then it would be very very very much appreciated! :)
2. A mixture of Cu(NO3)2 and Cu(NO3)2.2.5H20 has a mass of 3.150 g. Upon heating the water is driven off and the mass of the Cu(NO3)2 is 2.853 g. Determine the % by weight of the Cu(NO3)2.2.5H20 in the original mixture
(Use the isotopic mass of 19F = 18.9984 amu.)
2. I don't know what Cu(NO3)2.2.5H2O is... no subscripts
Cu(NO3)2. 2.5H2O is copper (II) nitrate hemipentahydrate, molar mass of 232.59g mol-1.
m(H2O) = 3.150 - 2.853
= 0.297
n(H2O) = 0.297/18.0
= 0.0165
H2O : Cu(NO3)2. 2.5H2O
2.5 : 1
Thus n(Cu(NO3)2. 2.5H2O) = 0.0165/2.5
= 0.006600
Thus m(Cu(NO3)2. 2.5H2O) = 0.006600 x 232.59
= 1.535094
Thus % w/w Cu(NO3)2. 2.5H2O = 1.535094/3.150
= 0.48733142857
Thus % w/w Cu(NO3)2. 2.5H2O = 48.73%