Equivalent mass of ${\mathrm{KMnO}}_4$ in acidic, basic, and neutral are in the ratio of

The correct option is D

3:15:5

Explanation:

Equivalent weight: It is the mass of one equivalent, that is the mass of a given substance that will combine with or displace a fixed quantity of another substance.

• The number of parts by mass of a substance that combines with or displaces 1.008 parts by mass of hydrogen or 8.0 parts of oxygen or 35.5 parts of chlorine is called its equivalent mass (EM).

Formula to calculate the equivalent mass

• Equivalent weight = Molar mass/number of electrons lost or gained (n)
• The molar mass of ${\mathrm{KMnO}}_4$=$\mathrm{atomic}\mathrm{mass}\mathrm{of}\mathrm{K}+\mathrm{atomic}\mathrm{mass}\mathrm{of}\mathrm{Mn}+(4\times \mathrm{atomic}\mathrm{mass}\mathrm{of}\mathrm{oxygen})=39.10+54.94+(4\times 16)=158.04\mathrm{g}/\mathrm{mol}$

In Acidic medium

$8{\mathrm{H}}^{+}+{{\mathrm{MnO}}_4}^{-}+5{\mathrm{e}}^{-}\to {\mathrm{Mn}}^{2+}+4{\mathrm{H}}_2\mathrm{O}$

• Here n = 5
• Eq wt = $\frac{158.04}{5}$
• Eq wt = 31.61g/equivalent

In Basic Medium

${{\mathrm{MnO}}_4}^{-}+{\mathrm{e}}^{-}\to {{\mathrm{MnO}}_4}^{2-}$

• Here n = 1
• Eq wt = 158.04/1
• Eq wt = 158.04g/Equivalent

In Neutral Medium

${{\mathrm{MnO}}_4}^{-}+4{\mathrm{H}}^{+}+3{\mathrm{e}}^{-}\to {\mathrm{MnO}}_2+2{\mathrm{H}}_2\mathrm{O}$

• Here n = 3
• Eq wt = $\frac{158.04}{3}$
• Eq wt = 52.68g/Equivalent

Therefore the 31.61g/equivqlent, 158.04g/Equivalent and 52.68g/Equivalent in the ratio of 3:15:5

Hence option (D) is the correct answer.