2. Sample Papers
  3. KVPY

KVPY Sample Paper KVPY Stream-SX Model Paper-29

  • question_answer
    A mixture of \[{{\operatorname{CuSO}}_{4}}.5{{H}_{2}}\operatorname{O}\] and \[{{\operatorname{MgSO}}_{4}}.7{{H}_{2}}O\] is heated until all the water is lost. If 5.020 g of the mixture gives 2.988 g of the anhydrous salts, what is the percent by mass of\[{{\operatorname{CuSO}}_{4}}.5{{H}_{2}}\operatorname{O}\] in the mixture?

    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:\[\frac{3.557g}{5.020g}\times 100%\]%
    \[=70.86%\]%


You need to login to perform this action.
You will be redirected in 3 sec spinner