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Molality : It is defined as the moles of the solute pressent in 1 kg of the solvent . It is denoted by m. Molality(m)`=("Number of moles of solute")/("Number of kilograms of the solvent")` let `w_(A)` grams of the solute of molecular mass `m_(A)` be present in `w_(B)` grams of the solvent, then: Molality(m) =`(w_(A))/(m_(A)xxw_(B))xx1000` Relation between mole fraction and molality: `X_(A)=(n)/(N+n) "and" X_(B)=(N)/(N+n)` `(X_(A))/(X_(B))=(n)/(N)=("Moles of solute")/("Moles of solvent")=(w_(A)xxm_(B))/(w_(B)xxm_(A))` `(X_(A)xx1000)/(X_(B)xxm_(B))=(w_(A)xx1000)/(w_(B)xxm_(A))= m or (X_(A)xx1000)/((1-X_(A))m_(B))=m` The molality of 1 litre solution with y% by (w/v) pf `CaCO_(3)` is 2 . The weight of the solvent present in the solution is 900g , then value of y is : [Atomic weight : Ca=40, C=12 , O=16]A. 9B. 18C. 27D. 36

Answer» Correct Answer - B


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