\(\therefore \) \(1\) molecule \({C_{60}}{H_{122}}\) has mass \(\frac{{842}}{{6 \times {{10}^{23}}}}\)
\( = 140.333 \times {10^{ - 23}}gm\) \( = 1.4 \times {10^{ - 21}}\,gm\).
$2MnO_4^ - + 5{C_2}O_4^ - + 16{H^ + } \to 2M{n^{ + + }} + 10C{O_2} + 8{H_2}O$
અહી $20\, mL$ of $0.1\, M\, KMnO_4$ એ કોના બરાબર હશે