JEE Mains Model Paper 13

Let the reaction be R (OH)n + 2CH3MgI → nCH4 + R(OMgI)n
Amount of gas produced =

1.34‌ml

22400‌ml

= 6 × 10−5 mole
Amount of compounded used =

1

n

(6 × 10−5)
This is equal to (1.79 × 10−3 g/90g/mole) = 2 × 10−5 mole
Hence n =

6×10−5‌mole

2×10−5‌mole

= 3