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Chemistry
The treatment of CH3OH with CH3MgI releases 1.04 mL of a gas at STP. The mass of CH3OH added is
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Q. The treatment of $CH_3OH$ with $CH_3MgI$ releases $1.04 \,mL$ of a gas at STP. The mass of $CH_3OH$ added is
Alcohols Phenols and Ethers
A
1.49 mg
46%
B
2.98 mg
24%
C
3.71 mg
19%
D
4.47 mg
10%
Solution:
$\,{}^nCH_3OH = \,{}^nCH_4$
$ = \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$