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JEE Main & Advanced JEE Main Paper (Held On 9 April 2014)

  • question_answer
    Dissolving 120 g of a compound of (mol. wt. 60) in 1000 g of water gave a solution of density 1.12 g/mL. The molarity of the solution is:   [JEE Main Online Paper ( Held On 09 Apirl  2014  )

    A) 1.00M                  

    B) 2.00 M

    C) 2.50 M                 

    D) 4.00 M

    Correct Answer: B

    Solution :

                    Given mass of solute (w) = 120 g mass of solvent (w) = 1000 g Mol. mass of solute = 60 g density of solution = 1.12 g/ ml From the given data, Mass of solution = 1000 + 120 = 1120 g \[\because \]\[d=\frac{Mol.\,mass}{v}\]or\[V=\frac{Mol.\,mass}{d}\] Volume of solution\[V=\frac{1120}{1.12}=1000ml\]or = 1 litre Now molarity \[(M)=\frac{W}{Mol.\,mass\times V(lit)}=\frac{120}{60\times 1}=2M\]


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