Question

If you have 3.48 x 1023 formula units of Lead (II) Permanganate, what is its mass?

Answer 1

Answer:

$\large \boxed{\text{257 g}}$

Explanation:

We can convert the formula units to moles and then use the molar mass to find the mass.

1. Moles of Pb(MnO₄)₂

$n = 3.48 \times 10^{23}\text{ formula units} \times \dfrac{\text{1 mol}}{6.022 \times 10^{23}\text{ formula units}} = \text{0.5779 mol}$

 

2. Mass of Pb(MnO₄)₂

$\text{Mass} = \text{0.5779 mol} \times \dfrac{\text{445.07 g}}{\text{1 mol u}} = \text{257 g}\\\\\text{The mass of lead(II) permanganate is \large \boxed{\textbf{257 g}} }$