JEE MAIN - Chemistry (2019 - 10th April Evening Slot - No. 3)
Which one of the following graphs between molar conductivity ($${\Lambda _m}$$) versus $$\sqrt C $$ is correct ?
_10th_April_Evening_Slot_en_3_2.png)
_10th_April_Evening_Slot_en_3_4.png)
_10th_April_Evening_Slot_en_3_6.png)
_10th_April_Evening_Slot_en_3_8.png)
Explanation
The graph is drawn using following equation,
$${\Lambda _m} = \Lambda _m^\infty - b\sqrt c $$
As the size of K+ is higher than the size of Na+, then the hydration radii of aqueous Na+ will be more than the aqueous K+. Therefore the ionic mobility of Na+ will be smaller than K+. Hence the conductance of K+ will be higher.
$$ \therefore $$ $$\Lambda _m^\infty $$ of K+ > $$\Lambda _m^\infty $$ of Na+
$$ \Rightarrow $$ $$\Lambda _m^\infty $$(KCl) > $$\Lambda _m^\infty $$(NaCl)
So, y intercept is more for KCl.
As slope b is constant both the lines will be parallel.
$${\Lambda _m} = \Lambda _m^\infty - b\sqrt c $$
As the size of K+ is higher than the size of Na+, then the hydration radii of aqueous Na+ will be more than the aqueous K+. Therefore the ionic mobility of Na+ will be smaller than K+. Hence the conductance of K+ will be higher.
$$ \therefore $$ $$\Lambda _m^\infty $$ of K+ > $$\Lambda _m^\infty $$ of Na+
$$ \Rightarrow $$ $$\Lambda _m^\infty $$(KCl) > $$\Lambda _m^\infty $$(NaCl)
So, y intercept is more for KCl.
As slope b is constant both the lines will be parallel.
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