Question

The number of moles in 3.00 g of boron tribromide $BB{r}_3$ is 'x' $\times {10}^{-2}$. 'x' is :

• A. $1.19$
• B. $2.39$
• C. $5.56$
• D. $2.19$

Answer 1

The correct option is A $1.19$

Molar mass of $BB{r}_3$ = 11 + 3(80)
= 251 g
Now 251 g of $BB{r}_3$ will contain 1 mole
So, 3 g will contain = $\frac{1}{251}\times 3$
= $1.19\times {10}^{-2}$ moles