The number of moles in 3.00 g of boron tribromide $B B r_{3}$ is 'x' $\times 10^{- 2}$ . 'x' is .

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Solution

Molar mass of $B B r_{3}$ = 11 + 3(80) = 251 g
Now 251 g of $B B r_{3}$ will contain 1 mole
So, 3 g will contain = $\frac{1}{251} \times 3$
= $1.19 \times 10^{- 2}$ moles

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The number of moles in 3.00 g of boron tribromide BBr3 is 'x' ×10−2. 'x' is .

Molar mass of BBr3 = 11 + 3(80) = 251 g Now 251 g of BBr3 will contain 1 mole So, 3 g will contain = 1251×3 = 1.19×10−2 moles

The number of moles in 3.00 g of boron tribromide $B B r_{3}$ is 'x' $\times 10^{- 2}$ . 'x' is .

Molar mass of $B B r_{3}$ = 11 + 3(80) = 251 g
Now 251 g of $B B r_{3}$ will contain 1 mole
So, 3 g will contain = $\frac{1}{251} \times 3$
= $1.19 \times 10^{- 2}$ moles