Arrange the following solutions in increasing order of their concentration.
(A) 7.5 g KCl (74.5 g $mol^{-1}$) in 100 mL solution
(B) 3 g Urea (60 g $mol^{-1}$) in 1 L solution
(C) 3.6 g glucose (180 g $mol^{-1}$) in 100 mL solution
(D) 4 g NaOH (40 g $mol^{-1}$) in 1 L solution
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (B), (D), (C), (A)
To arrange the solutions in increasing order of concentration, we calculate the molarity (M) of each solution.
Molarity (M) = (mass / molar mass) / volume in litres
|
Solution |
Mass |
Molar mass |
Volume |
Moles |
Molarity (M) |
|
(A) KCl |
7.5 g |
74.5 g/mol |
100 mL = 0.1 L |
7.5/74.5 = 0.10067 |
1.0067 M |
|
(B) Urea |
3 g |
60 g/mol |
1 L |
3/60 = 0.05 |
0.05 M |
|
(C) Glucose |
3.6 g |
180 g/mol |
100 mL = 0.1 L |
3.6/180 = 0.02 |
0.20 M |
|
(D) NaOH |
4 g |
40 g/mol |
1 L |
4/40 = 0.1 |
0.10 M |
Molarity values:
- (B) Urea → 0.05 M
- (D) NaOH → 0.10 M
- (C) Glucose → 0.20 M
- (A) KCl → 1.0067 M
Increasing order of concentration (molarity): (B) < (D) < (C) < (A)