# Question 5c1f1

Aug 20, 2017

$157.48 \text{ g}$

#### Explanation:

Given:

$\frac{1.45 \times {10}^{24} \text{ atoms of Zn}}{1}$

Multiply by the conversion factor for the Avogadro constant :

$\left(1.45 \times {10}^{24} \text{ atoms of Zn")/1(1" mole of Zn")/(6.02xx10^23" atoms of Zn}\right)$

Please observe how the conversion factor cancels the units "atoms of Zn" and we are left with the units "moles of Zn":

$\left(1.45 \times {10}^{24} \cancel{\text{ atoms of Zn"))/1(1" mole of Zn")/(6.02xx10^23cancel(" atoms of Zn}}\right)$

Multiply by the conversion factor for the Molar Mass of Zn :

(1.45xx10^24cancel(" atoms of Zn"))/1(1" mole of Zn")/(6.02xx10^23cancel(" atoms of Zn"))(65.38" g of Zn")/(1" mole of Zn")

Please observe that the units "mole of Zn" cancel and we are left with the units " g of Zn":

$\left(1.45 \times {10}^{24} \cancel{\text{ atoms of Zn"))/1(1cancel(" mole of Zn"))/(6.02xx10^23cancel(" atoms of Zn"))(65.38" g of Zn")/(1cancel(" mole of Zn}}\right)$

All that is left to do is the multiplication and division and we know that we have correct units:

(1.45xx10^24)/1(1)/(6.02xx10^23)(65.38" g of Zn")/(1) = 157.48" g of Zn"#