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Solutions and Colloids: MCQs and Answers with explanation

1. 18g glucose is dissolved in 180g of water. The relative lowering of vapour pressure is





... Answer is c)
Explanation:


2. 100g of a 10% (w/w) NaOH contains 10g of NaOH in.





... Answer is d)
Explanation: First we need to calculate the mass of solute and solvent from 10%(W/W) NaOH.
10% (W/W) NaOH contains 10 gram (w) of NaOH(solute) dissolved in 100g (w) of solution.
So, mass of solute = 10g
Mass of solution = 100g

3. Which of the following W/W solutions has the lowest freezing point?





... Answer is c)
Explanation: Depression in freezing point is colligative property and it is directly related to molality of solute. Solution with highest molality will show highest freezing point depression and hence lowest freezing point.
In 18% (W/W) glucose
So, mass of glucose (C6H12O6) = 18g
No. of moles of glucose = 18/180 = 0.1 moles
Mass of solution = 100g
Mass of solvent = Mass of solution - Mass of solute
Mass of solvent = 100 - 18
Mass of solvent = 82g = (82/1000) kg = 0.082 Kg
In 6% (W/W) Urea (NH2CONH2)
So, mass of urea = 6g
No. of moles of urea = 6/60 = 0.1 moles
Mass of solution = 100g
Mass of solvent = 100 - 6
Mass of solvent = 94g = (94/1000) kg = 0.094 Kg
In 34.2% (W/W) Sucrose (C12H22O1)
So, mass of sucrose = 34.2g
No. of moles of glucose = 34.2/342 = 0.1 moles
Mass of solution = 100g
Mass of solvent = 100 - 34.2
Mass of solvent = 65.8g = (65.8/1000) kg = 0.0658 Kg
In 9 % (W/W) glucose (C6H12O6)
So, mass of glucose = 9g
No. of moles of glucose = 9/180 = 0.05 moles
Mass of solution = 100g
Mass of solvent = 100 - 9
Mass of solvent = 91g = (91/1000) kg = 0.091 Kg
Hence molality of 34.2% of sucrose is highest, so it will show lowest depression in freezing point.

4. All solutions containing 1g of non-volatile solute will have





... Answer is d)
Explanation: 1g of different solute has different number of solute particles depending on their molar masses. Due to different number of solute particles, physical properties such as vapour pressure, boiling point and freezing point also different for each solution . So, correct option is d.

5. Molarity of pure water is





... Answer is c)
Explanation: Density of water = 1kg/dm3.
Density of water = 1000g/dm3
Molar mass of water = 18gmol-1
No. of moles / dm3 = (1000/18)/dm3
No. of moles / dm3 = 55.5 moles/dm3
Thus 1dm3 contains 55.5 moles

6. A solution of urea (Mol. wt. = 60) is 10% (W/V), the volume in which 1 mole of it is dissolved will be





... Answer is b)
Explanation: 10% (W/V) urea (NH2CONH2) contains 10 gram (w) of urea (solute) dissolved in 100 cm3 (v) of solution.
So, No. of moles of urea = 10 g/60 gmol-
No. of moles of urea = 0.16 moles
Volume of urea solution = 100 cm/
Volume of urea solution = 100/1000 dm/ = 0.1 dm/
0.167 moles of urea = 0.1 dm/
1 moles of urea = 0.1 dm// 0.167
1 moles of urea = 0.6 moles

7. The solubility of a substance decreases with increase in temperature if heat of solution is





... Answer is b)
Explanation: Positive heat of solution is for endothermic process and negative heat of solution is for exothermic process. So, if we increase the temperature, exothermic process is disfavoured which will result in decrease in solubility of substance in solution.

8. Molarity of glucose solution when 9g of it is dissolved in 250 cm3 of solution is





... Answer is b)
Explanation:Mass of glucose (C6H12O6) = 9g and volume of solution = 250 cm3
No. of moles of glucose = 9 g/ 60gmol- = 0.05 moles
Volume of solution = 250/1000 dm3 = 0.25dm3
Molarity of glucose solution = No. of moles / Volume of solution
Molarity of glucose solution = 0.05 mole/ 0.25 dm3
Molarity of glucose solution = 0.2 M

9. Sea water has about 6 ppm dissolved oxygen. What mass of dissolved oxygen is present in one Kg of sea water





... Answer is c)
Explanation: Sea water with 6 ppm dissolved oxygen has 6 parts (by weight or volume) of oxygen per million parts (by weight or volume) of sea water.
Let
106 g of sea water contains = 6 g of oxygen
1 g of sea water will contain = 6 g/106
1000g (1kg) of sea water will contain = 1000 x 6 g/106
= 6 x 10-3 g = 6 x 10-6.Kg

10. Which of the following is not true for a colloid?





... Answer is c)
Explanation: Colloid cannot pas through the ultra filter paper due to larger particles of colloid particle size than pores of ultra filter paper. However, colloid particles can pass through an ordinary filter paper. Some important MCQs

11. Name the partially miscible liquids from the following:





... Answer is b)
Explanation: A pair of liquids which donot completely mix with with each other and form two layer over a range of temperature is called partially miscible liquids. Nicotin and water are partially miscible liquids due to hydrogen bonding beween nitrogen of nicotine and hydrogen from water.

12. Solubility of sodium chloride in water is possible because :





... Answer is a)
Explanation: When NaCl is mixed with water, partial negatively charged oxygen atoms of water molecules start to surround (solvate) positively charged sodium ions and partial positvely charged hydrogen atoms of water molecules start to surround the negatively charged chroide ions. During solvation, new forces of attractions are created due to which energy is released known as hydration energy. Hydration energy is greater that lattic energy of NaCl. Therefore sodium chloride is dissolved when mixed with water. If lattic energy of any salt is greater than its hydration energy, that salt will not dissolve.

13. Solution contains 85.5g of sucrose(C6H12O6) in 250cm3. What is its molarity?





... Answer is a)
Explanation: Mass of sucrose (C12H22O11) = 85.5g and volume of solution = 250 cm3
No. of moles of sucrose = 85.5 g/ 342gmol- = 0.25 moles
Volume of solution = 250/1000 dm3 = 0.25dm3
Molarity of sucrose solution = No. of moles / Volume of solution
Molarity of sucrose solution = 0.25 mole/ 0.25 dm3
Molarity of glucose solution = 1 M

14. Roults law is represented by:





... Answer is d)
Explanation: According to Roult law
P = PO.X1 ...(i) option a)
where P= Vapour pressure of solution, PO = Vapour pressure of pure solvent, X1 = Mole fraction of pure solvent.
Equations in option b and c can be derived by substituting the value of X1 in equation (i),
As sum of mole fractions is unity,
X1+X2 = 1 or
X1 = 1-X2 Now, equaiton i can be written as
P = PO(1-X2) = PO -POX2
P - PO = -POX2
PO - P = POX2
∆P = PO.X2 -----(ii) option b)
( or
∆P / PO =X2 -----(iii) option c)

15. 18g of glucose is dissolved in 90g of water. The relative lowering of vapour pressure is equal to ?





... Answer is c)
Explanation: Mass of glucose (C6H12O6) = 18g and mass of water (solvent) = 90g
No. of moles of glucose = 18 g/ 180gmol- = 0.1 moles
No. of moles of water = 90 g/ 18gmol- = 5 moles
mole fraction of glucose = No. of moles of glucose/ (No. of moles of glucose + No. of moles of water)
mole fraction of glucose = 0.1/ (5 + 0.1) = 1/5.1 =10/51
By definition of Roult law,
The relative lowering of vapour pressue (∆P / PO) = mole fraction of solute (X2) = 10/51 = 10/51
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