When 0.01 mol of an organic compound containing 60% carbon was burnt completely, 4.4g of CO2 was produced. The molar mass of compound is g mol−1 (Nearest integer)

Book Online Demo

Check Your IQ

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

When 0.01 mol of an organic compound containing 60% carbon was burnt completely, 4.4g of $C O_{2}$ was produced. The molar mass of compound is $g m o l^{- 1}$ (Nearest integer)

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

answer is 200.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Let M is the molar mass of the compound (g/mol)
mass of compound = 0.01M gm.
Mass of carbon = 0.01M $\times \frac{60}{100}$
Moles of $C O_{2}$ from combustion $= \frac{4.4}{44} = m o l e s o f$ carbon
$\frac{0.01 M}{12} \times \frac{60}{100} = \frac{4.4}{44}$
$M = \frac{4.4}{44} \times \frac{100}{60} \times \frac{12}{0.01} = 200 g m / m o l$