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# MCQ MOLARITY - MOLALITY - MOLE FRACTION - PERCENT COMPOSITION IIT JEE - AIEEE

## 1) Two solutions of a substance (non electrolyte) are mixed in the following manner. 480 mL of 1.5 M first solution + 520 mL of 1.2 M second solution. What is the molarity of the final mixture?

(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'.

## Related questions

### 2) The density (in g mL–1) of a 3.60 M sulphuric acid solution that is 29% H2SO4 (Molar mass = 98 g mol–1) by mass will be:

(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'.

### 3) 25 mL of a solution of barium hydroxide on titration with 0.1 molar solution of hydrochloric acid gave titer value of 35 mL. The molarity of barium hydroxide solution was :

(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'.

### 4) The density of a solution prepared by dissolving 120 g of urea (mol.mass = 60 u) in 1000 g of water is 1.15 g/mL. The molarity of this solution is:

(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'.

### 5) Density of a 2.05 M solution of acetic acid in water is 1.02 g/mL. The molality of the solution is:

(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'.

### 6) With increase in temperature, which of these changes?

(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.

### 7) A 5.2 molal aqueous solution of methyl alcohol, CH3OH is supplied. What is the mole fraction of methyl alcohol in the solution?

(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'.

### Homework

1) Define molarity.

2) Define molality.

3) What is mole fraction?

4) Which concentration term has no units?

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