JEE Main & Advanced Sample Paper JEE Main Sample Paper-4

  • question_answer
    A solution contains a mixture of \[N{{a}_{2}}C{{O}_{3}}\] and NaOH. Using phenolphthalein as indicator, 25 mL of mixture required 19.5 mL of 0.995 N HCI for the end point. Whith methyl organge, 25 mL of solution required 24 mL of the same HCI for the end point. The grams per litre of \[N{{a}_{2}}C{{O}_{3}}\] in the mixture is

    A)  23.2                      

    B)  18.5    

    C)  19.9                      

    D)  12.8

    Correct Answer: A

    Solution :

    Let the moles of \[N{{a}_{2}}C{{O}_{3}}\] and NaOH in 25 mL mixture be x and y respectively. Case -1: When HPh is used as indicator \[\underset{x}{\mathop{NaOH}}\,+\underset{x}{\mathop{HCl\to }}\,NaCl+{{H}_{2}}O\] \[\underset{y}{\mathop{N{{a}_{2}}C{{O}_{3}}}}\,+\underset{y}{\mathop{HCl\to }}\,NaHC{{O}_{3}}+NaCl\] So\[,\]\[x+y=19.5\times {{10}^{-3}}\times (0.995\times 1)\]        ?.1 Case - 2 : When MeOH is used as indicator. \[\underset{x}{\mathop{NaOH}}\,+\underset{x}{\mathop{HCl\to }}\,NaCl+{{H}_{2}}O\] \[\underset{y}{\mathop{N{{a}_{2}}C{{O}_{3}}}}\,+\underset{2y}{\mathop{2HCl\to 2}}\,NaCl+{{H}_{2}}O+C{{O}_{2}}\] .....(ii)  On solving equation (i) and (ii), we get, \[x=13.93\times {{10}^{-3}}mol\] and \[y=5.4725\times {{10}^{-3}}mol\] Now, weight of NaOH in 25 mL mixture \[=13.93\times {{10}^{-3}}\times 40=557.2\times {{10}^{-3}}g\] Weight of \[N{{a}_{2}}C{{O}_{3}}\] in 25 mL mixture \[=5.4725\times {{10}^{-3}}\times 106=580.085\times {{10}^{-3}}g\] \[\therefore \] Weight of NaOH per litre \[=\frac{557.2\times {{10}^{-3}}\times 1000}{25}=23.2034g/L\] weight of \[N{{a}_{2}}C{{O}_{3}}\]per litre \[=\frac{580.085\times {{10}^{-3}}\times 1000}{25}=23.2034g/L\]

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