ROH + CH_3MgI → ROMgI + CH_4(g) moles of CH_4 = moles of ROH ⇒ V2 2 4 0 0= mM.M(Assuming NTP Condition) ⇒ 3.12 2 4 0 0= 4.5×1 0^−3M.M ⇒M M=3 2.5 1 Nearest Integer = 33

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Correct answer: 33. (33) ROH + CH_3MgI → ROMgI + CH_4(g) moles of CH_4 = moles of ROH ⇒ V2 2 4 0 0= mM.M (Assuming NTP Condition) ⇒ 3.12 2 4 0 0= 4.5×1 0^−3M.M ⇒M M=3 2.5 1 Nearest Integer = 33

$ROH + C H_{3} MgI \to ROMgI + C H_{4} (g) moles of C H_{4} = moles of ROH \Rightarrow \frac{V}{22400} = \frac{m}{M . M} (Assuming NTP Condition) \Rightarrow \frac{3.1}{22400} = \frac{4.5 \times 1 0 ^{- 3}}{M . M} \Rightarrow M M = 32.51 Nearest Integer = 33$

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