## Chemistry (7th Edition)

The molarity of glucose in that solution is equal to $5.0\times 10^{- 3}$ mol/L.
$100$ mL = $100 \times 10^{-3}$ L = 0.1 L $90$ mg = $90 \times 10^{-3}$ g = 0.09 g 1. Calculate the molar mass $(C_6H_{12}O_6)$: 12.01* 6 + 1.008* 12 + 16* 6 = 180.156g/mol 2. Calculate the number of moles $(C_6H_{12}O_6)$ $n(moles) = \frac{mass(g)}{mm(g/mol)}$ $n(moles) = \frac{ 0.09}{ 180.156}$ $n(moles) = 5.0\times 10^{- 4}$ 3. Find the concentration in mol/L $(C_6H_12O_6)$: $C(mol/L) = \frac{n(moles)}{volume(L)}$ $C(mol/L) = \frac{ 5.0\times 10^{- 4}}{ 0.1}$ $C(mol/L) = 5.0\times 10^{- 3}$