Answer
Approximately 0.1029M.
Work Step by Step
1. Find the molar mass of $Na_2CrO_4:$
Na: 23 * 2 = 46
Cr: 52 * 1 = 52
O: 16 * 4 = 64
46 + 52 + 64 = 162g/mol
2. Find the number of mols in 12.5g of $Na_2CrO_4$:
$mm = \frac{mass(g)}{n(moles)} $
$162 = \frac{12.5}{n(moles)}$
$n(moles) = \frac{12.5}{162}$
$n(moles) \approx 0.07716$
3. Now, find the concentration:
$C = \frac{n(moles)}{V(L)}$
$C = \frac{0.07716}{0.750}$
$C \approx 0.10288M $
