(AIEEE 2005 - equivalent to IIT JEE MAIN)
1) 1.20 M
2) 1.344 M
3) 1.50 M
4) 2.70 M
Logic & Solution:
Molarity of a mixture, Mmix can be calculated by:
Mmix = (M1V1 + M2V2) / (V1 + V2) = (1.5 x 480) + (1.2 x 520)/(480 + 520) = 720+624/1000 = 1.344 M
Conclusion:
The correct option is '2'.
(AIEEE 2007)
1) 1.45
2) 1.64
3) 1.88
4) 1.22
Solution:
The relation between Molarity, M and mass percent (%) is given by:
M = (% x 10 x d) / MW
Where:
MW = molecular weight of solute
d = density of solution
Therefore:
d = M x MW / (% x 10) = 3.60 x 98 / 29 x 10 = 1.216 g mL-1
Conclusion:
The correct option is '4'.
(AIEEE 2003)
1) 0.07
2) 0.14
3) 0.28
4) 0.35
Solution:
The reaction involved in the titration is:
Ba(OH)2 + 2HCl -------> BaCl2 + 2H2O
The equation that is used to find the molarity of unknown solutions in titrations is:
M1V1/n1 = M2V2/n2
Where
M1 and M2 are, respectively, the molarities of Ba(OH)2 & HCl solutions.
V1 and V2 are, respectively, the volumes of Ba(OH)2 & HCl solutions.
n1 = Stoichiometric coefficient of Ba(OH)2 = 1
n2 = Stoichiometric coefficient of HCl = 2
Therefore:
molarity of barium hydroxide solution, M1 = (M2V2/n2) x (n1/V1) = (0.1 x 35 / 2) x (1/25) = 0.07 M
Conclusion:
The correct option is '1'.
(AIEEE 2012)
1) 1.02 M
2) 2.05 M
3) 0.50 M
4) 1.78 M
Solution:
Molarity of a solution (M) = n/V
Where:
n = number of moles of urea = 120/60 = 2 mol
V = volume of the solution = mass of solution / density of solution = (120 + 1000) g / 1.15 g mL-1 = 974 mL = 0.974 L
Therefore:
M = 2 mol / 0.974 L = 2.05 mol L-1
Note: Volume of solution should be expressed in litres.
Conclusion:
The correct option is '2'.
(AIEEE 2006)
1) 1.14 mol kg-1
2) 3.28 mol kg-1
3) 2.28 mol kg-1
4) 0.44 mol kg-1
Solution:
Molarity of acetic acid solution = 2.05 M
Therefore the number of moles of acetic acid in 1 L solution = 2.05 mol
mass of acetic acid in 1 L solution = no. of moles x molar mass = 2.05 mol x 60 g mol-1 = 123 g
Note: molar mass of acetic acid is 60 g mol-1
Density, d = mass of solution / volume of solution
Hence:
mass of solution = density x volume of solution = 1.02 g mL-1 x 1000 mL = 1020 g
mass of solvent = mass of solution - mass of solute = 1020 - 123 = 897 g = 0.897 kg
molality, m = no. of moles / mass of solvent (in Kg) = 2.05 mol / 0.897 kg = 2.285 mol kg-1
Conclusion:
The correct option is '3'.
(AIEEE 2002)
1) molality
2) fraction of solute present in water
3) weight fraction of solute
4) mole fraction
Logic:
Temperature will affect the volume of solution. Hence the expression for concentration that contain volume will change with temperature.
Conclusion:
The weight fraction of solute present in water will change with temperature.
(AIEEE 2011)
(1) 0.050
(2) 1.100
(3) 0.190
(4) 0.086
Logic & solution:
molality of aqueous solution = 5.2 mol kg-1
Therefore, the number of moles of methyl alcohol in 1 kg of solvent (water) = 5.2 mol
And the number of moles of solvent (water) = weight / molar mass = 1000 g / 18 g mol-1 = 55.55 mol
Hence:
Mole fraction of methyl alcohol = nMeOH /nMeOH + nwater = 5.2 / 55.55 + 5.2 = 0.08559
Conclusion:
The correct option is '4'.
1) Define molarity.
2) Define molality.
3) What is mole fraction?
4) Which concentration term has no units?
Author: Aditya vardhan Vutturi