## Chemistry (7th Edition)

$500$ mL = $500 \times 10^{-3}$ L = 0.500 L 1. Calculate the molar mass $(KCl)$: 39.1* 1 + 35.45* 1 = 74.55g/mol 2. Calculate the number of moles $(KCl)$ $n(moles) = \frac{mass(g)}{mm(g/mol)}$ $n(moles) = \frac{ 37.3}{ 74.55}$ $n(moles) = 0.500$ 3. Find the concentration in mol/L $(KCl)$: $C(mol/L) = \frac{n(moles)}{volume(L)}$ $C(mol/L) = \frac{ 0.500}{ 0.500}$ $C(mol/L) = 1.00$