Test Index

NCERT Class XII Chemistry
Chapter - Solutions
Questions with Solutions

© examsnet.com
Question : 3 of 53
Marks: +1, -0
Define the following terms :
(i) Mole fraction
(ii) Molality
(iii) Molarity
(iv) Mass percentage.
Solution:  
(i) Mole fraction : Mole fraction of a constituent (solute as well as solvent) is the ratio of the number of moles of one component to the total number of moles of all the components present in the solution. If nAn_A and nBn_B are the number ofmoles of solvent and solute respectively then
xA=nAnA+nBx_{A}= \frac{n_{A}}{n_{A}+n_{B}} and xB=nBnA+nBx_{B}= \frac{n_{B}}{n_{A}+n_{B}}
(ii) Molality : Molality of a solution is defined as the number of moles of the solute dissolved in one kilogram of the solvent. It is denoted by m. It is expressed as
m=Number of moles of soluteMass of solvent in kgm= \frac{\text{Number of moles of solute}}{\text{Mass of solvent in kg}} or Number of moles of soluteMass of solvent in grams ×1000\frac{\text{Number of moles of solute}}{\text{Mass of solvent in grams }} \times 1000
(iii)Molarity : Molarity of a solution is defined as the number of moles of the solutedissolved in one litre or one dm3^{3} of the solution. It is denoted by M and is expressed as
M=Number of moles of solute Volume of solution in litresM= \frac{\text{Number of moles of solute }}{\text{Volume of solution in litres}}
(iv) Mass percentage : It may be defined as mass of solute in grams per 100 g of solution.
wW\frac{w}{W} or mass percentage =Mass of soluteTotal mass of solution×100= \frac{\text{Mass of solute}}{\text{Total mass of solution}} \times 100
© examsnet.com
Go to Question: