# What is the molar quantity of a 458*g mass of "lead(II) sulfide"?

Jun 1, 2017

Approx. $2 \cdot m o l$.

#### Explanation:

The molar mass of $\text{lead sulfide}$ is $239.3 \cdot g \cdot m o {l}^{-} 1$.

And to get the molar quantity of the given mass of lead sulfide, we take the quotient.

$\text{Moles of lead sulfide"="Mass"/"Molar mass} = \frac{458 \cdot g}{239.3 \cdot g \cdot m o {l}^{-} 1} =$

$1.91 \cdot m o l$.

This is dimensionally consistent because $\frac{458 \cdot g}{239.3 \cdot g \cdot m o {l}^{-} 1} =$

$\frac{\cancel{g}}{\cancel{g} \cdot m o {l}^{-} 1} = \frac{1}{\frac{1}{m o l}} = m o l$ as required..............