JEE Main & Advanced JEE Main Paper Phase-I (Held on 08-1-2020 Morning)

  • question_answer
    The stoichiometry and solubility product of a salt with the solubility curve given below is, respectively [JEE MAIN Held On 08-01-2020 Morning]

    A) \[{{X}_{2}}Y,\text{ }2\times {{10}^{9}}\text{ }{{M}^{3}}\]

    B) \[X{{Y}_{2}},\text{ }4\times {{10}^{9}}\text{ }{{M}^{3}}\]

    C) \[X{{Y}_{2}},\text{ }1\times {{10}^{9}}\text{ }{{M}^{3}}\]

    D) \[XY,2\times {{10}^{6}}\text{ }{{M}^{3}}\]

    Correct Answer: B

    Solution :

    [b] From the given curve, if \[\left[ X \right]=1\text{ }mM\text{ }then\left[ Y \right]=2\text{ }mM\] \[\therefore \]Salt is \[X{{Y}_{2}}\] \[{{K}_{sp}}=\left[ X \right]{{\left[ Y \right]}^{2}}=\left( {{10}^{3}} \right){{\left( 2\times {{10}^{3}} \right)}^{2}}=4\times {{10}^{9}}\text{ }{{M}^{3}}\]


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