Question

The loss in mass for $1\text{mole}$ of potassium cation in the salt $K_{2}S{O}_4$ will be:
($K_{2}S{O}_4\to 2K^{+}+S{O}_4^{2-}$)

• A. $5.48\times {10}^{-7}\text{kg}$
• B. $7.48\times {10}^{-9}\text{kg}$
• C. $4.50\times {10}^{-5}\text{kg}$
• D. $30.96\times {10}^{-7}\text{kg}$

Answer 1

The correct option is A $5.48\times {10}^{-7}\text{kg}$

As we know,
$K_{2}S{O}_4\to 2K^{+}+S{O}_4^{2-}$

From the above reaction, we can say that in the salt $K_{2}S{O}_4$, potassium cation has $+1$ unit positive charge.

$\Rightarrow$ charge of $+1$ unit developed due to loss of $1e^{-}$ per cation $K^{+}$.

In one mole of $K^{+}$ cation, total number of electrons lost is,

$n_e=N_A=6.023\times {10}^{23}$

Thus total loss in mass for $1\text{mole}$ of cation.
$\Delta m=n_e{m}_e$
$\Rightarrow \Delta m=(6.023\times {10}^{23})\times (9.1\times {10}^{-31}\text{kg})$

$\therefore \Delta m=5.48\times {10}^{-7}\text{kg}$

Hence, option $\left(a\right)$ is correct.

$\begin{array}{c}\text{Why this question ?}\\ \text{Caution: Charge on potassium cation is +1 unit in salt}{K}_{2}S{O}_4\\ \text{and loss in mass of potassium cation is asked for 1 mole}\\ \text{of cation and "must not be confused with 1 mole of salt".}\end{array}$