What is the mass of 12.04xx10^23 "formula units" of MgCl_2?

Jul 14, 2017

Well, we know that $6.022 \times {10}^{23}$ $M g C {l}_{2}$ formula units have a mass of $95.21 \cdot g$.......

Explanation:

Do we know this? All I have is taken the molar mass of magnesium chloride, and translated this into a NUMBER of magnesium chloride formula units...

And here we gots $12.04 \times {10}^{23}$ $M g C {l}_{2}$ formula units.........

And thus we gots the product...........

$\left(12.04 \times {10}^{23} \cdot \text{formula units")/(6.022xx10^23*"formula units} \cdot m o {l}^{-} 1\right) \times 95.21 \cdot g \cdot m o {l}^{-} 1$....

And I make this a mass of $190.42 \cdot g$........do you agree?

Is the product dimensionally consistent?