## Chemistry: Molecular Approach (4th Edition)

The concentration of $NaNO_3$ is equal to 0.214 M.
1. Calculate the molar mass $(NaNO_3)$: 22.99* 1 + 14.01* 1 + 16* 3 = 85g/mol 2. Calculate the number of moles $(NaNO_3)$ $n(moles) = \frac{mass(g)}{mm(g/mol)}$ $n(moles) = \frac{ 45.4}{ 85}$ $n(moles) = 0.5341$ 3. Find the concentration in mol/L $(NaNO_3)$: $C(mol/L) = \frac{n(moles)}{volume(L)}$ $C(mol/L) = \frac{ 0.5341}{ 2.5}$ $C(mol/L) = 0.214$