JEE Main & Advanced Sample Paper JEE Main - Mock Test - 44

  • question_answer
    \[30\text{ }mL\]of a solution containing \[9.15g\,{{L}^{-1}}\] of an oxalate \[{{K}_{x}}{{H}_{y}}{{({{C}_{2}}{{O}_{4}})}_{z}}.n{{H}_{2}}O\]required for titration \[27\text{ }mL\]of \[0.12\text{ }N\text{ }NaOH\]and \[36\text{ }mL\]of \[0.12\text{ }N\]\[KMn{{O}_{4}}\] for oxidation. Find the ratio \[x:y:z\]and the value of n.

    A) \[1:3:2\]and \[2\] 

    B)        \[2:1:3\]and \[1\]

    C) \[3:1:2\]and \[3\]     

    D)        \[2:3:1\]and \[1\]

    Correct Answer: A

    Solution :

    [a] \[{{K}_{x}}{{H}_{y}}{{({{C}_{2}}{{O}_{4}})}_{z}}.n{{H}_{2}}O\] Normality of oxalate as an acid \[=\frac{27\times 0.12}{30}=0.108\] Ew of oxalate as acid             \[=\frac{Strength}{N}=\frac{9.15}{0.108}=84.72\] Normality of oxalate as reducing agent                         \[=\frac{36\times 0.12}{30}=0.144\] Ew of oxalte as reducing agent \[\frac{9.15}{0.144}=63.54\] Oxalate as acid: \[\frac{Mw}{y}=84.72\] Oxalate as reducing agent \[\frac{Mw}{2z}=63.54\] \[x+y=2z\] \[y=1.5z\] \[\therefore \,\,\,\,\,\,\,x:y:z=1:3:2\] \[\therefore \,\,\,\,\,\,\,39\times 1+3\times 1+2\times 88+18n=254\] \[n=2\] Formula is \[K{{H}_{3}}{{({{C}_{2}}{{O}_{4}})}_{2}}.2{{H}_{2}}O\]


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