JEE Main 2019 — Electrochemistry Question with Solution

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.