# How many molecules are in 720 g of C_6H_12O_6?

May 14, 2018

Approx. $36 \times {10}^{23} \cdot \text{sugar molecules}$

#### Explanation:

We work out (i) the molar quantity of sugar present...

${n}_{\text{sugar}} = \frac{720 \cdot g}{180 \cdot g \cdot m o {l}^{-} 1} = 6 \cdot m o l$

But there are $6.022 \times {10}^{23} \cdot \text{particles} \cdot m o {l}^{-} 1$...

...and so (ii) we take the product...

$6.022 \times {10}^{23} \cdot \text{particles"*mol^-1xx6*mol=?*"particles}$