Answer
The concentration of $Na_2CO_3$ is equal to 0.254M.
Work Step by Step
1. Calculate the molar mass $(Na_2CO_3)$:
22.99* 2 + 12.01* 1 + 16* 3 = 105.99g/mol
2. Calculate the number of moles $(Na_2CO_3)$
$n(moles) = \frac{mass(g)}{mm(g/mol)}$
$n(moles) = \frac{ 6.73}{ 105.99}$
$n(moles) = 0.0635$
3. Find the concentration in mol/L $(Na_2CO_3)$:
1000 ml = 1 L
250 ml = 0.250 L
$C(mol/L) = \frac{n(moles)}{volume(L)}$
$ C(mol/L) = \frac{ 0.0635}{ 0.250} $
$C(mol/L) = 0.254$
