Define the following terms `:`
`a.` Mole fraction `b.` Molality
`c.` Molarity `d. Mass percentage.

Explanation

(i) Molar fraction: it is defined as the ratio of the number of moles of the solute to the total number of moles in the solution. If A is the number of moles of solute dissolved in B moles of solvent, then mole fraction of solute.
`(X_(A))+(n_(A))/(n_(A)+n_(B))` . . .(1)
Mole fration of solvent `(X_(B))=(n_(B))/(n_(A)+n_(B))` . . .(2)
Adding the above two equations, we get
`X_(A)+X_(B)=(n_(A))/(n_(A)+n_(B))+(n_(B))/(n_(A)+n_(B))=(n_(A)+n_(B))/(n_(A)+n_(B))=1`
i.e., `X_(A)+x_(B)=1`
`thereforeX_(A)=1-X_(B)` or `X_(B)=1-X_(A)`
(ii). Molality: It is defined as die number of moles of a solute present in 1000g (1kg) of a solvent
Molality `(m)=("Number of moles of solute")/("weight of solvent in kg")=(n)/(W)`
Note: Molality is considered better way of expressing concentration of solutions, as compared to molarity because molality does not changewith change in temperature since the mass of solvent does not vary with temperature.
(iii). Molarity: it is defined as the number of moles of solute present in one litre of solution.
Molarity(M)=
`("Number of moles of solute")/("Volume of solution in litre")=(n)/(V)`
`n=("Weight in grams")/("Molecular weight of solute")`
`thereforeM=("Weight in grams")/("Volume of solution in litre")`
`xx(1)/("Molecular weight of solute")`
Strength: this is weight (in gms) of solute per litre of solution
`therefore"Molarity"=("Strength")/("Molecular weight of Solute")`
or Strength=Molarity `xx` Molecular weight
Note: Molarity is the most common way of expressing concentration of a solution in lobaratory. However, it has one disadvantage. it chagnes with temperature because volume of a solution alters due to expansion and contraction of the liquid with temperature.
(iv). Mass percentage: It is the amount of solute in grams present in 100 g of solution
`=("mass of solute")/"Mass of solution")xx100`