## Chemistry (7th Edition)

The molarity of the solution is equal to $0.0684$ $M$.
$100$ mL = $100 \times 10^{-3}$ L = 0.100 L $400$ mg = $400 \times 10^{-3}$ g = 0.400 g 1. Calculate the molar mass $(NaCl)$: 22.99* 1 + 35.45* 1 = 58.44g/mol 2. Calculate the number of moles $(NaCl)$ $n(moles) = \frac{mass(g)}{mm(g/mol)}$ $n(moles) = \frac{ 0.400}{ 58.44}$ $n(moles) = 6.84\times 10^{- 3}$ 3. Find the concentration in mol/L $(NaCl)$: $C(mol/L) = \frac{n(moles)}{volume(L)}$ $C(mol/L) = \frac{ 6.84\times 10^{- 3}}{ 0.100}$ $C(mol/L) = 0.0684$ $M$