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Answer :
Given, Mass of the sample compound = 0.24g, mass of boron = 0.096g, mass of oxygen = 0.144g
\( Percentage \ of \ boron = \) \( mass \ of \ boron \over mass \ of \ the \ compound \) \( × 100 \) \( = \) \( 0.096 \over 0.24 \) × 100 = 40%
Percentage of oxygen = 100 – Percentage of boron
= 100 – 40 = 60%