The solution having same ionic strength as 0.01 m ZnCl2 is,
a) 0.01 m KCl,
b) 0.02 m KCl,
c) 0.03 m KCl,
b) 0.025 m KCl,
Answers
Explanation:
1.MELON
2.BANANA
3.PEAR
4.LEMON
5.FIG
6.ORANGE
7.APPLE
8.CHERRY
9.PAPAYA
10.PLUM
11.GUAVA
12.PEACH
13.LYCHEE
14. KIWI
Concept:
The ionic strength can be calculated by the formula which calculates the sum of each ion’s molar concentration multiplied by the valence squared.
Ionic strength, I =(1/2) ∑ⁿ₁ (CₓZₓ²)
where C is the concentration and Z is the charge on each ion.
Given:
The solution of 0.01 M ZnCl₂.
Find:
The ionic strength of 0.01 M ZnCl₂.
Solution:
Ionic Strength of 0.01 M ZnCl₂,
I =(1/2) ∑ⁿ₁ (CₓZˣ²) = (1/2) [(0.01×2²) +(0.01×1²)]
I=(1/2) (0.04+0.01) = 0.05/2 = 0.025
Calculating the ionic strength of given options,
Ionic Strength of 0.01 M KCl,
I =(1/2) ∑ⁿ₁ (CₓZˣ²) = (1/2) [(0.01×1²) +(0.01×1²)]
I=(1/2) (0.01+0.01) = 0.02/2 = 0.01
Ionic Strength of 0.02 M KCl,
I =(1/2) ∑ⁿ₁ (CₓZˣ²) = (1/2) [(0.02×1²) +(0.02×1²)]
I=(1/2) (0.02+0.02) = 0.04/2 = 0.02
Ionic Strength of 0.03 M KCl,
I =(1/2) ∑ⁿ₁ (CₓZˣ²) = (1/2) [(0.03×1²) +(0.03×1²)]
I=(1/2) (0.03+0.03) = 0.03/2 = 0.03
Ionic Strength of 0.025 M KCl,
I =(1/2) ∑ⁿ₁ (CₓZˣ²) = (1/2) [(0.025×1²) +(0.025×1²)]
I=(1/2) (0.025+0.025) = 0.05/2 = 0.025
The Ionic Strength of 0.025 M KCl is 0.025 and the Ionic Strength of 0.01 M ZnCl₂ is also 0.025. So, the Ionic Strength of 0.025 M KCl and the Ionic Strength of 0.01 M ZnCl₂ are the same.
Hence, The solution 0.025 M KCl will have the same ionic strength as 0.01 M ZnCl₂. Hence, the correct option is (d) 0.025 M KCl.
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