A 0.24 , g sample of a compound of oxygen and | vedclass

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

Question

(A) Mass of the compound $= 0.24 \, g$Mass of boron $= 0.096 \, g$Mass of oxygen $= 0.144 \, g$Percentage of boron $= \frac{	ext{Mass of boron}}{	ex

Vedclass · Accepted Answer

$O = 60\%, B = 40\%$

Vedclass · Answer

(A) Mass of the compound $= 0.24 \, g$Mass of boron $= 0.096 \, g$Mass of oxygen $= 0.144 \, g$Percentage of boron $= \frac{	ext{Mass of boron}}{	ext{Mass of compound}} 	imes 100$$= \frac{0.096 \, g}{0.24 \, g} 	imes 100 = 40\%$Percentage of oxygen $= \frac{	ext{Mass of oxygen}}{	ext{Mass of compound}} 	imes 100$$= \frac{0.144 \, g}{0.24 \, g} 	imes 100 = 60\%$Alternative method:Percentage of oxygen $= 100\% - 	ext{Percentage of boron}$$= 100\% - 40\% = 60\%$

$A$ $0.24 \, g$ sample of a compound of oxygen and boron was found by analysis to contain $0.096 \, g$ of boron and $0.144 \, g$ of oxygen. Calculate the percentage composition of the compound by weight.

Similar Questions

Which postulate of Dalton's atomic theory can explain the law of definite proportions?

Calculate the molecular masses of $H_2$,$O_2$ and $Cl_2$.

Which has more number of atoms,$100\,g$ of sodium or $100\,g$ of iron (given,atomic mass of $Na = 23\,u, Fe = 56\,u$)?

Define the atomic mass unit.

Convert the following into moles:
$(a)$ $12\, g$ of oxygen gas
$(b)$ $20\, g$ of water
$(c)$ $22\, g$ of carbon dioxide.

Vedclass Test Series

Exam Paper Generator

Online Exam Module