Question

1. An experiment requires a solution that is 80% methyl alcohol by mass. What mass of methyl alcohol should be added to 20 g of water to make this solution?

2. Calculate the mole fraction of NaCl in a solution containing 12.6 g NaCl and 21.3 g KCl in 122 g of water.

3.A solution of phosphoric acid was made by dissolving 10.0 g of H3PO4 in 100.0 mL of water. The resulting volume was 104 mL. Calculate the molarity of the solution. Assume water has a density of 1.00 g/cm3.

Answer 1

1) $\omega = \mathrm{m}$ (substance) / $\mathrm{m}$ (solution) * 100%

m (solution) = m (substance) + m (solvent)

$\omega = \mathrm{m}$ (substance) / $\mathrm{m}$ (substance) + $\mathrm{m}$ (solvent) * 100%

m (solvent) = m (H₂O) = 20 g

$\omega = 80\%$ or 0.8

m (substance) = m (methyl alcohol) = X

$$0.8 = \mathrm{X} / (\mathrm{X} + 20)$$ $$0.8 * \mathrm{X} + 16 = \mathrm{X}$$ $$16 = 0.2 * \mathrm{X}$$ $$\mathrm{X} = 80$$

m (substance) = m (methyl alcohol) = 80 g

2) $n_{\mathrm{NaCl}}$ – amount of NaCl = $m_{\mathrm{NaCl}} / \mathrm{M}(\mathrm{NaCl}) = 12.6 \, \mathrm{g} / 58.5 \, \mathrm{g} / \mathrm{mole} = 0.215$ moles

$n_{\mathrm{KCl}}$ – amount of KCl = $m_{\mathrm{KCl}} / \mathrm{M}(\mathrm{KCl}) = 21.3 \, \mathrm{g} / 74.5 \, \mathrm{g} / \mathrm{mole} = 0.2859$ moles

$n_{\mathrm{H_2O}}$ – amount of $\mathrm{H_2O} = \mathrm{m}_{\mathrm{H_2O}} / \mathrm{M}(\mathrm{H_2O}) = 122 \, \mathrm{g} / 18 \, \mathrm{g} / \mathrm{mole} = 6.778$ moles

N – mole fraction of NaCl

$n_{\mathrm{tot}}$ – total amount of all constituents in a solution = $n_{\mathrm{NaCl}} + n_{\mathrm{KCl}} + n_{\mathrm{H_2O}} = 0.215 \, \mathrm{moles} + 0.2859 \, \mathrm{moles} + 6.778 \, \mathrm{moles} = 7.2789 \, \mathrm{moles}$

$\mathrm{N} = \mathrm{n}_{\mathrm{NaCl}} / (\mathrm{n}_{\mathrm{NaCl}} + \mathrm{n}_{\mathrm{KCl}} + \mathrm{n}_{\mathrm{H_2O}}) = 0.215 \, \mathrm{moles} / 7.2789 \, \mathrm{moles} = 0.0295$ or $2.95\%$

3) $\mathrm{M}(\mathrm{H_3PO_4}) = 98 \, \mathrm{g} / \mathrm{mole}$

$n = \mathrm{m}(\mathrm{H_3PO_4}) / \mathrm{M}(\mathrm{H_3PO_4}) = 10 \, \mathrm{g} / 98 \, \mathrm{g} / \mathrm{mole} = 0.102 \, \mathrm{moles}$

V (solution) = 104 mL = 0.104 L

$C_M = n$ (solute) / V (solution, L) = 0.102 moles / 0.104 L = 0.98 M