# How many atoms of Boron are present in 35.76 g of Boron?

Nov 15, 2015

Molar mass of $B$ $=$ $10.81$ $g \cdot m o {l}^{-} 1$. If there are 36 odd grams of $B$, there are approx. $\frac{7}{2}$ $\times$ ${N}_{A}$ boron atoms, where ${N}_{A}$ is Avogadro's number.

#### Explanation:

$\frac{35.76 \cdot g}{10.81 \cdot g \cdot m o {l}^{-} 1}$ $\times$ $6.022 \times {10}^{23}$ $m o {l}^{-} 1$ $=$ ??

Nov 15, 2015

$1.991 \times {10}^{24}$

#### Explanation:

Molar mass of B = 10.81g/mol
So, 1 mole of B = 10.81g
and 1 mole = ${N}_{A}$= $6.023 \times {10}^{23}$
By combining,
10.81g of B = $6.023 \times {10}^{23}$ atoms of B
thus,
35.76g of B = $\frac{6.023 \times {10}^{23}}{10.81} \times 35.76$
35.76g of B = $1.991 \times {10}^{24}$