a.
\[\begin{aligned}n_\mathrm{NaCl} &= \frac {m_\mathrm{NaCl}}{M_\mathrm{NaCl}} = \frac {0,50\mathrm{g}}{(22,99 + 35,45)\mathrm{g/mol}} = 0,00855578\mathrm{mol} \\ c_\mathrm{NaCl} &= \frac {n_\mathrm{NaCl}}{V} = \frac {0,00855578\mathrm{mol}}{1\mathrm{dm^3}} = 0,00855578\mathrm{mol/dm^3} ≈ 0,0086\mathrm{M}\end{aligned}\]
b.
\[\begin{aligned} n_\mathrm{C_2H_5OH} &= \frac {m_\mathrm{C_2H_5OH}}{M_\mathrm{C_2H_5OH}} = \frac {0,125\mathrm{g}}{(12,01 \cdot 2 + 1,008 \cdot 6 + 16,00)\mathrm{g/mol}} = 0,00271338\mathrm{mol} \\ c_\mathrm{C_2H_5OH} &= \frac {n_\mathrm{C_2H_5OH}}{V} = \frac {0,00271338\mathrm{mol}}{0,250\mathrm{dm^3}} = 0,01085352\mathrm{mol/dm^3} ≈ 0,0109\mathrm{M}\end{aligned}\]
c.
\[\begin{aligned} n_\mathrm{CH_3COOH} &= \frac {m_\mathrm{CH_3COOH}}{M_\mathrm{CH_3COOH}} = \frac {0,33\mathrm{g}}{(12,01 \cdot 2 + 1,008 \cdot 4 + 16,00 \cdot 2)\mathrm{g/mol}} =\\&= 0,00549524\mathrm{mol} \\ c_\mathrm{CH_3COOH} &= \frac {n_\mathrm{CH_3COOH}}{V} = \frac {0,00549524\mathrm{mol}}{0,650\mathrm{dm^3}} = 0,00845421\mathrm{mol/dm^3} ≈ 0,0085\mathrm{M}\end{aligned}\]
d.
\[\begin{aligned} n_\mathrm{C_6H_{12}O_6} &= \frac {m_\mathrm{C_6H_{12}O_6}}{M_\mathrm{C_6H_{12}O_6}} = \frac {0,045\mathrm{g}}{(12,01 \cdot 6 + 1,008 \cdot 12 + 16,00 \cdot 6)\mathrm{g/mol}} =\\ &= 0,00024978\mathrm{mol} \\ c_\mathrm{C_6H_{12}O_6} &= \frac {n_\mathrm{C_6H_{12}O_6}}{V} = \frac {0,00024978\mathrm{mol}}{0,016\mathrm{dm^3}} = 0,01561147\mathrm{mol/dm^3} ≈ 0,016\mathrm{M}\end{aligned}\]
