The number of moles of solute present in `1kg` of solvent is called

The number of moles of the solute present in 600 ml of 0.5M solution is:

The number of moles of the solute present in 500 ml of 0.5M solution is:

The number of moles of the solute present in 500 ml of 0.06M solution is:

The number of moles of the solute present in 500 ml of 0.4M solution is:

The number of moles of the solute present in 600 ml of 0.6M solution is:

The number of moles of the solute present in 500 ml of 0.6M solution 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 If the mole fraction of a solute is changed from (1)/(4) "to" (1)/(2) in the 800 g of solvent then the ratio tof molality will be:

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 mole fraction of the solute in the 12 molal solution of CaCo_(3) is :

The number of moles of the solute present in 600 ml of 0.05 M solution is: