JEE Main 2026 — Jan 24, Shift 2 — Chemistry solution

JEE Main2026Jan 24, Shift 2Chemistry

Grignard reagent $RMgBr$ (P) reacts with water and forms a gas (Q). One gram of Q occupies 1.4 $dm^{3}$ at STP. (P) on reaction with dry ice in dry ether followed by $H_{3} O^{+}$ forms a compound (Z). 0.1 mole of (Z) will weigh ________ g. (Nearest integer)

Your answer

Solution

1 g of Q occupies 1.4 dm $^{3}$ at STP $\Rightarrow$ molar volume $\frac{22.4}{1.4} = 16$ g/mol, so Q is $CH_{4}$ .

Therefore the Grignard reagent is $CH_{3} MgBr$ .

With dry ice ( $CO_{2}$ ) followed by $H_{3} O^{+}$ :

$CH_{3} MgBr (i) CO_{2} (dry ice) (ii) H_{3} O^{+} CH_{3} COOH$

Z = acetic acid, molar mass = 60 g/mol.

Mass of 0.1 mole $= 0.1 \times 60 = 6$ g.

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