Question

If A for Ba(OH)2, BaCl2 and NH4CI are x, y and z respectively then A of NH40H is

Answer 1

Answer:

The reaction between (barium hydroxide) Ba(OH)2 + NH4Cl (ammonium chloride). This reaction is a double displacement gas evolution reaction. The products of this reaction are Barium Chloride(BaCl2), Ammonia (NH3) and Water (H2O).

Answer 2

The correct answer is $z+\frac{x}{2} -\frac{y}{2}$. (Option b)

Given:

Lamda for $BaOH_{2}$, $BaCl_{2}$ and $NH_{4}Cl$ is x, y and z respectively.

To Find:

Lamda of $NH_{4}OH$

Solution:

We can simply solve this problem by using the following process.

Let lamda be denoted by A.

Then,

As per Kohler's Law which states that "the equivalent conductivity of an electrolyte at infinite dilution is equal to the sum of the conductances of the anions and cations."

2A ($NH_{4}OH$) = 2A ($NH_{4}Cl$) + A ($BaOH_{2}$) - A ($BaCl_{2}$)

A ($NH_{4}OH$) = $\frac{2A(NH_{4}Cl)+A(BaOH_{2})-A(BaCl_{2})}{2}$

A ($NH_{4}OH$) = $\frac{2z+x-y}{2}$

A ($NH_{4}OH$) = $z+\frac{x}{2} -\frac{y}{2}$

Hence, the correct answer is $z+\frac{x}{2} -\frac{y}{2}$.

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