[" 9: The molality of a "1L" solution with "x%H_(2)SO_(4)" is 9.The weight of solvent present in the solution is 910grams "],[" The value of "x" is: "]

The molality of a 1 L solution with x % H_2SO_4 is 9. The weight of solvent present in the solution is 910 grams. The value of x is :

The molality of 1 L solution with x% H_(2)SO_(4) is equal to 9. The weight of the solvent present in the solution is 910 g . The value of x is:

The molality of 1 L solution with x% H_(2)SO_(4) is equal to 9. The weight of the solvent present in the solution is 910 g . The value of x is:

The molality of 1 L solution with x% H_(2)SO_(4) is equal to 9. The weight of the solvent present in the solution is 910 g . The value of x in g per 100 mL is

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]

The molality of X% H_(2) SO_(4) solution is equal to 9. The weight of the solvent present in the solution is 910g.The value of 'X' is : 90, 80.3, 40.13, 9

The molality of an H_(2)SO_(4) solution is 9. The weight of the solute in 1 kg H_(2)SO_(4) solution is: