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JEE Main & Advanced JEE Main Paper (Held on 08-4-2019 Morning)

  • question_answer
    100 mL of a water sample contains 0.81 g of calcium bicarbonate and 0.73 of magnesium bicarbonate. The hardness of this water sample expressed in terms of equivalents of \[CaC{{O}_{3}}\]is: (molar mass of calcium bicarbonate is 162 g \[mo{{l}^{-1}}\]and magnesium bicarbonate is 146 \[gmo{{l}^{-1}}\])                                                                                       [JEE Main 8-4-2019 Morning]

    A) 1,000 ppm                  

    B) 10,000 ppm

    C) 100 ppm                     

    D) 5,000 ppm

    Correct Answer: B

    Solution :

    \[{{n}_{eq.}}CaC{{O}_{3}}={{n}_{eq}}Ca{{(HC{{O}_{3}})}_{2}}+{{n}_{eq}}Mg{{(HC{{O}_{3}})}_{2}}\]           or,\[\frac{W}{100}\times 2=\frac{0.81}{162}\times 2+\frac{0.73}{146}\times 2\]           \[\therefore \]\[w=1.0\]           \[\therefore \]Hardness \[=\frac{1.0}{100}\times {{10}^{6}}=10000ppm\] Correct option : (b)


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