Question

How many grams of sodium chlorate(nacio3) are in 1.23 x10^23 representative particles?

Answer 1

About 21.7 grams.

Step-by-step explanation:

To find the mass of a sample of sodium chlorate (NaClO₃) given the amount of particles present, we can convert from particles to moles, and moles to mass using its molecular weight.

The molecular weight of sodium chlorate is:

`\displaystyle \begin{aligned} \mathcal{M}_\text{NaClO$_3$} &= (22.99 + 35.45 + 3(16.00)) \text{ g/mol} \\ \\ & = 106.44\text{ g/mol} \end{aligned}`

And recall that in one mole of any substance, there are 6.02 × 10²³ of it. Hence:

`\displaystyle \begin{aligned} 1.23* 10^(23) \text{ NaClO$_3$} &\cdot \frac{1\text{ mol NaClO$_3$}}{6.02* 10^(23)\text{ NaClO$_3$}} \cdot \frac{106.44\text{ g NaClO$_3$}}{1\text{ mol NaClO$_3$}} \\ \\ &= 21.7 \text{ g NaClO$_3$} \end{aligned}`

In conclusion, there are about 21.7 grams of sodium chlorate.