Question bafc0

Dec 14, 2015

$0 , 0017 m o l$

Explanation:

Since concentration is moles per unit volume ($c = \frac{n}{V}$), and moles is mass over molar mass (n=m/(M_r#), we may combine the two to obtain that

$c = \frac{m}{{M}_{r} \cdot V}$

$\therefore m = c \cdot {M}_{r} \cdot V$

$= 2 , 28 m o l / {\mathrm{dm}}^{3} \times \left(39 + 80\right) g / m o l \times 0 , 00075 {\mathrm{dm}}^{3}$

$= 0 , 2035 g$.

This corresponds to the following number of moles :

$n = \frac{m}{{M}_{r}} = \frac{0 , 2035 g}{119 g / m o l} = 0 , 0017 m o l$.