0.01 mol of an organic compound (X) containi | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Moles of $H_2O$ produced $= \frac{0.9 \ g}{18 \ g \ mol^{-1}} = 0.05 \ mol$. Moles of $H$ atoms $= 0.05 \ mol 	imes 2 = 0.1 \ mol$. Mass of $H$ a

Vedclass · Accepted Answer

$100$

Vedclass · Answer

(A) Moles of $H_2O$ produced $= \frac{0.9 \ g}{18 \ g \ mol^{-1}} = 0.05 \ mol$. Moles of $H$ atoms $= 0.05 \ mol 	imes 2 = 0.1 \ mol$. Mass of $H$ atoms in $0.01 \ mol$ of $(X) = 0.1 \ mol 	imes 1 \ g \ mol^{-1} = 0.1 \ g$. Mass of $H$ atoms in $1 \ mol$ of $(X) = \frac{0.1 \ g}{0.01 \ mol} = 10 \ g \ mol^{-1}$. Given that the compound contains $10\%$ hydrogen by mass,we have: $10\% = \frac{	ext{Mass of } H 	ext{ in } 1 \ mol 	ext{ of } (X)}{	ext{Molar mass of } (X)} 	imes 100$. $10 = \frac{10}{M} 	imes 100$. $M = 100 \ g \ mol^{-1}$.

$0.01 \ mol$ of an organic compound $(X)$ containing $10\%$ hydrogen,on complete combustion produced $0.9 \ g \ H_2O$. Molar mass of $(X)$ is $.......... \ g \ mol^{-1}$.

Similar Questions

The empirical formula of a compound is $C_2H_5O$ and its vapour density is $45$. What is the molecular formula of the compound?

An organic compound on analysis gives the following data: $C = 57.82 \%$,$H = 3.6 \%$ and the rest is oxygen. Its vapour density is $83$. Calculate its molecular formula.

The empirical formula of an acid is $CH_2O_2$. The molecular formula of this acid will be .....

The empirical formula of a crystalline iron compound is $Fe_2(SO_4)_3$. It is used in water and sewage treatment to remove suspended impurities. The percentage composition of iron,sulfur,and oxygen in this compound is respectively......

$A$ compound of carbon,hydrogen,and nitrogen contains three elements in the respective ratio of $9 : 1 : 3.5$ grams. The empirical formula for the compound is

Vedclass Test Series

Exam Paper Generator

Online Exam Module