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JEE Main & Advanced JEE Main Paper (Held On 25 April 2013)

  • question_answer
                    10 mL of 2 (M) of \[\operatorname{NaOH}\] solution is added to 200 mL of 0.5 (M) of NaOH solution. What in the final concentration?              JEE Main Online Paper ( Held On 25  April 2013 )       

    A)                 0.57 (M)                              

    B)                                        5.7 (M)

    C)                                        11.4 (M)                              

    D)                                        1.14 (M)              

    Correct Answer: A

    Solution :

                    From molarity equation \[{{M}_{1}}{{V}_{1}}+{{M}_{2}}{{V}_{2}}=M{{V}_{(total)}}\] \[2\times \frac{10}{1000}+0.5\times \frac{200}{1000}=M\times \frac{210}{1000}\] \[120=M\times 210\] \[M=\frac{120}{210}=0.57M\]


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