IIT JEE - AIEEE

(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, M_{mix} can be calculated by:

M_{mix} = (M_{1}V_{1} + M_{2}V_{2}) / (V_{1} + V_{2}) = (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 -------> BaCl_{2} + 2H_{2}O

The equation that is used to find the molarity of unknown solutions in titrations is:

M_{1}V_{1}/n_{1} = M_{2}V_{2}/n_{2}

Where

M_{1} and M_{2} are, respectively, the molarities of Ba(OH)_{2} & HCl solutions.

V_{1} and V_{2} are, respectively, the volumes of Ba(OH)_{2} & HCl solutions.

n_{1} = Stoichiometric coefficient of Ba(OH)_{2} = 1

n_{2} = Stoichiometric coefficient of HCl = 2

Therefore:

molarity of barium hydroxide solution, M_{1} = (M_{2}V_{2}/n_{2}) x (n_{1}/V_{1}) = (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 = n_{MeOH} /n_{MeOH} + n_{water} = 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