Some Basic Concepts of Chemistry - Result Question 85
####23. The equivalent weight of $MnSO _4$ is half of its molecular weight, when it converts to
$(1988,1 M)$
(a) $Mn _2 O _3$
(b) $MnO _2$
(c) $MnO _4^{-}$
(d) $MnO _4^{2-}$
Objective Question II (More than one correct option)
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Solution:
- Mass of an electron $=9.108 \times 10^{-31} kg$
$\because 9.108 \times 10^{-31} kg=1.0$ electron
$\therefore \quad 1 kg=\frac{1}{9.108 \times 10^{-31}}$ electrons $=\frac{10^{31}}{9.108} \times \frac{1}{6.023 \times 10^{23}}$
$$ =\frac{1}{9.108 \times 6.023} \times 10^{8} \text { mole of electrons } $$
