The treatment of CH3OH with CH3MgI releases 1.04 mL of a gas at STP. The mass of CH3OH added is

Question

The treatment of CH3OH with CH3MgI releases 1.04 mL of a gas at STP. The mass of CH3OH added is (A) 1.49 mg (B) 2.98 mg (C) 3.71 mg (D) 4.47 mg

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/4cf89eb03fb8c1121e140df281a26b3e-.png" /> nCH3OH = nCH4 = (1.04/22400) = 4.64 × 10-5 4.64 × 10-5 = ( textmass of CH3OH/ textmolar mass ) = (x/32) x (mass of CH3OH) = 32 × 4.64 × 10-5 = 1.49 × 10-3 g = 1.49 mg

Q. The treatment of $C H_{3} O H$ with $C H_{3} M g I$ releases $1.04 m L$ of a gas at STP. The mass of $C H_{3} O H$ added is

1.49 mg

2.98 mg

3.71 mg

4.47 mg

$^{n} C H_{3} O H =^{n} C H_{4}$
$= \frac{1.04}{22400} = 4.64 \times 1 0^{- 5}$
$4.64 \times 1 0^{- 5} = \frac{mass of C H _{3} O H}{molar mass} = \frac{x}{32}$
$x$ (mass of $C H_{3} O H$ )
$= 32 \times 4.64 \times 1 0^{- 5}$
$= 1.49 \times 1 0^{- 3} g$
$= 1.49 m g$

Q. The treatment of CH3​OH with CH3​MgI releases 1.04mL of a gas at STP. The mass of CH3​OH added is