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]

Options

A. 9
B. 18
C. 27
D. 36

Answer

B