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JEE Main & Advanced JEE Main Paper (Held On 10-Jan-2019 Morning)

  • question_answer
    A mixture of 100 m mol of \[Ca{{\left( OH \right)}_{2}}\], and 2 g of sodium sulphate was dissolved in water and the volume was made up to 100 mL. The mass of calcium sulphate formed and the concentration of \[O{{H}^{-}}\]in resulting solution, respectively, are: (Molar mass \[Ca{{\left( OH \right)}_{2}}\], \[N{{a}_{2}}S{{O}_{4}}\], and \[CaS{{O}_{4}}\] are 74, 143 and \[136\text{ }g\text{ }mo{{l}^{-1}}\], respectively; \[{{K}_{sp}}\]of  \[Ca{{\left( OH \right)}_{2}}\]is \[5.5\,\times \,{{10}^{-6}}\]) [JEE Main Online Paper (Held On 10-Jan-2019 Morning]

    A) \[13.6g,\text{ }0.28\text{ }mol\text{ }{{L}^{-1}}\]                    

    B)                  \[13.6\,g,\text{ }0.14\,mol\,{{L}^{-\,1}}\]

    C)                  \[1.9\,g,\text{ }0.28\text{ }mol\text{ }{{L}^{-1}}\]                    

    D)                  \[1.9\,g,\text{ }0.14\text{ }mol\,{{L}^{-1}}\]

    Correct Answer: C

    Solution :

    \[Ca{{\left( OH \right)}_{2}}\text{+}N{{a}_{2}}S{{O}_{4}}\to CaS{{O}_{4}}+2NaOH\] 100 m mole       14 m mole 86 m mole         - 14 m mole       28 m mole \[\therefore \text{ }W\text{ }CaS{{O}_{4}}=14\times {{10}^{-3}}\times 136\] \[=\text{ }1.9\text{ }gm\] \[[O{{H}^{-}}]=\frac{28}{100}\,\,=\,\,\,0.28\,M\]


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