# Chemistry Prelimary HELP (1 Viewer)

1. 2.

#### CM_Tutor

##### Moderator
Moderator
$\bg_white n(\text{NaI}) = \cfrac{m}{M} = \cfrac{8.85}{23.0+126.9} = k\ \text{mol}$

$\bg_white \text{From equation}: \quad \cfrac{n(\text{I}_2)}{n(\text{NaI})} = \cfrac{1}{2}$

$\bg_white n(\text{I}_2) = \cfrac{1}{2} \times k = \cfrac{k}{2}\ \text{mol}$

$\bg_white m(\text{I}_2) = nM = \cfrac{k}{2} \times (126.9 \times 2) = \cfrac{126.9 \times 8.85}{149.9}\ \text{g}$

#### CM_Tutor

##### Moderator
Moderator
$\bg_white n(\text{sodium acetate})\ \text{originally in solution}\ = CV = 1.55 \times 81 \times 10^{-3}\ \text{mol}$

$\bg_white n(\text{sodium acetate})\ \text{added as solid}\ = \cfrac{m}{M} = \cfrac{3.20}{82.0}\ \text{mol}$

$\bg_white n(\text{sodium acetate})\ \text{in final solution}\ = \left(1.55 \times 81 \times 10^{-3} + \cfrac{3.20}{82.0}\right)\ \text{mol}$

and assuming that all the added solid sodium acetate will dissolve (which it will), and that the volume of the solution does not change (which it actually will, though not by much), then the final solution will have

$\bg_white [\text{sodium acetate}]\ \text{in final solution}\ = \cfrac{n}{V} = \cfrac{1.55 \times 81 \times 10^{-3} + \cfrac{3.20}{82.0}}{81 \times 10^{-3}}\ \text{mol L}^{-1}$