Question

The treatment of $CH_3OH$ with $CH_3MgI$ releases $1.04 \,mL$ of a gas at STP. The mass of $CH_3OH$ added is

• A. 1.49 mg
• B. 2.98 mg
• C. 3.71 mg
• D. 4.47 mg

Answer 1

Answer: (A)

1.49 mg

Answer 2

$\,^nCH_3OH = \,^nCH_4$ $= \frac{1.04}{22400} = 4.64 \times 10^{-5}$ $4.64 \times 10^{-5} = \frac{\text{mass of } CH_3OH}{\text{molar mass }} = \frac{x}{32}$ $x$ (mass of $CH_3OH$) $= 32 \times 4.64 \times 10^{-5}$ $= 1.49 \times 10^{-3} \,g$ $= 1.49 \,mg$