A) 65.86
B) 70.86
C) 75.45
D) 79.25
Correct Answer: B
Solution :
The mass of water lost upon heating the mixture is \[\left( 5.020 g-2.988 g \right) = 2.032 g\]water. |
If\[\operatorname{x}=mass\,of\,CuS{{O}_{4}}.5{{H}_{2}}O\], then the mass of \[{{\operatorname{MgSO}}_{4}}7{{H}_{2}}O\,is\,(5.020-x)g.\]We can calculate the amount of water lost by each salt based on the mass % of water in each hydrate. |
We can write: |
\[(mass\,\,CuS{{O}_{4}}.5{{H}_{2}}O)\](%\[{{H}_{2}}O\])+\[(mass\,MgS{{O}_{4}}.7{{H}_{2}}O)\](%\[{{H}_{2}}O\]) |
= total mass \[{{H}_{2}}O=2.032g\,{{H}_{2}}O\] |
Calculate the \[%{{H}_{2}}O\]in each hydrate. |
%\[{{H}_{2}}O=\left( CuS{{O}_{4}}.5{{H}_{2}}O \right)\] |
\[=\frac{\left( 5 \right)\left( 18.02g \right)}{249.7g}\times 100%\]% |
= 36.08%\[{{H}_{2}}O\] |
%\[{{H}_{2}}O=\left( MgS{{O}_{4}}.7{{H}_{2}}O \right)\] |
\[=\frac{\left( 7 \right)\left( 18.02g \right)}{246.5g}\times 100%\]% |
= 51.17%\[{{H}_{2}}O\] |
Substituting into the equation above |
\[(x)(0.3608)+(5.020- x)(0.5117)= 2.032 g\] |
\[0.1509\text{ }x=0.5367\] |
\[x=3.557g=mass of\,CuS{{O}_{4}}. 5{{H}_{2}}O\] |
Finally, the percent by mass of\[CuS{{O}_{4}}\]. |
\[5{{H}_{2}}O\]In the mixture is:\[\frac{3.557g}{5.020g}\times 100%\]% |
\[=70.86%\]% |
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