將 \(7.1\ \text{g}\) 硫酸鈉固體溶解在水中，製備成 \(\text{250}\text{.0}\ \text{c}{{\text{m}}^{\text{3}}}\) 的硫酸鈉溶液。計算所得溶液的摩爾濃度。（答案保留至小數點後第二位。相對原子質量： \(\text{O}\ \text{=}\ \text{16}\text{.0}\)；\(\text{Na}\ \text{=}\ \text{23}\text{.0}\)；\(\text{S}\ \text{= 32}\text{.1}\)）
計算標準溶液的摩爾濃度要知道溶液中溶質的摩爾數和溶液的體積。
\[溶液的摩爾濃度\left( \text{mol}\ \text{d}{{\text{m}}^{-\text{3}}} \right)\ \text{=}\ \frac{\text{溶質的摩爾數}\left( \text{mol} \right)}{\text{溶液的體積}\left( \text{d}{{\text{m}}^{\text{3}}} \right)}\]\(\text{N}{{\text{a}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\ \text{的摩爾質量}\ \text{=}\ \text{23}\text{.0}\ \text{ }\!\!\times\!\!\text{ }\ \text{2}\ \text{+}\ \text{32}\text{.1}\ \text{+}\ \text{16}\text{.0}\ \text{ }\!\!\times\!\!\text{ }\ \text{4}\ \text{=}\ \text{142}\text{.1}\ \text{g}\ \text{mo}{{\text{l}}^{-1}}\)
\(\displaystyle{\text{N}{{\text{a}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\ \text{的摩爾數}\ =\ \frac{7.1\ \text{g}}{142.1\ \text{g}\ \text{mo}{{\text{l}}^{-1}}}\ =\ 0.050\ \text{mol}}\)
\(\displaystyle{\text{N}{{\text{a}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\ \text{溶液的摩爾濃度}\ \text{=}\ \frac{\text{0}\text{.050}\ \text{mol}}{\text{250}\text{.0}\ \text{c}{{\text{m}}^{\text{3}}}}\ \text{=}\ \frac{\text{0}\text{.050}\ \text{mol}}{\text{0}\text{.250}\ \text{d}{{\text{m}}^{\text{3}}}}\ \text{=}\ \text{0}\text{.20}\ \text{mol}\ \text{d}{{\text{m}}^{-3}}}\)
所以，所得溶液的摩爾濃度為 \(\text{0}\text{.20}\ \text{mol}\ \text{d}{{\text{m}}^{-3}}\)。
實驗室中要製備 \(\text{500}\text{.0}\ \text{mL}\) 濃度為 \(\text{0}\text{.1}\ \text{mol}\ \text{d}{{\text{m}}^{-3}}\) 的乙二酸（\({{\left( \text{COOH} \right)}_{\text{2}}}\)）標準溶液。計算所需稱量乙二酸晶體（\({{\left( \text{COOH} \right)}_{\text{2}}}\centerdot 2{{\text{H}}_{\text{2}}}\text{O}\)）的質量。（答案保留至小數點後第一位。乙酸晶體的摩爾質量為 \(\text{126}\text{.0}\ \text{g}\ \text{mo}{{\text{l}}^{-1}}\)。）
與練習一類似，要利用溶液摩爾濃度的公式，但計算順序相反。需要先計算溶液中溶質的摩爾數，再求出溶質的質量。
\[溶液的摩爾濃度\left( \text{mol}\ \text{d}{{\text{m}}^{-\text{3}}} \right)\ \text{=}\ \frac{\text{溶質的摩爾數}\left( \text{mol} \right)}{\text{溶液的體積}\left( \text{d}{{\text{m}}^{\text{3}}} \right)}\]\(\begin{align} \text{溶質的摩爾數}\ &= \ \text{500}\text{.0}\ \text{mL}\ \text{ }\!\!\times\!\!\text{ }\ \text{0}\text{.1}\ \text{mol}\ \text{d}{{\text{m}}^{-3}}\ \\ &= \ \text{0}\text{.500}\ \text{d}{{\text{m}}^{\text{3}}}\ \text{ }\!\!\times\!\!\text{ }\ \text{0}\text{.1}\ \text{mol}\ \text{d}{{\text{m}}^{-3}}\ \text{=}\ \text{0}\text{.05}\ \text{mol} \\ \end{align}\)
\(\text{溶質的質量}\ \text{=}\ \text{0}\text{.05}\ \text{mol}\ \text{ }\!\!\times\!\!\text{ }\ \text{126}\text{.0}\ \text{g}\ \text{mo}{{\text{l}}^{-1}}\ \text{=}\ \text{6}\text{.3}\ \text{g}\)
所以，所需乙二酸晶體的質量為 \(\text{6}\text{.3}\ \text{g}\)。
實驗室中購買的氫氯酸一般是質量百分比為 \(\text{36%}\)，密度為 \(\text{1}\text{.18 g}\ \text{c}{{\text{m}}^{\text{3}}}\) 的濃溶液。若將 \(\text{1}\text{.0 c}{{\text{m}}^{\text{3}}}\) 的原溶液稀釋成 \(\text{500}\text{.0}\ \text{c}{{\text{m}}^{\text{3}}}\) 的稀溶液，計算稀釋後溶液的濃度。(答案保留至小數點後第三位。氫氯酸的摩爾質量為 \(\text{36}\text{.5}\ \text{g}\ \text{mo}{{\text{l}}^{-1}}\)。）
稀釋前後溶質的質量或摩爾數相等。所以只要求出稀釋前溶質的摩爾數再除以稀釋後溶液的體積即可。
\(\text{1}\text{.0}\ \text{c}{{\text{m}}^{\text{3}}}\ \text{原溶液中含溶質的質量}\ \text{=}\ \text{1}\text{.0}\ \text{c}{{\text{m}}^{\text{3}}}\ \text{ }\!\!\times\!\!\text{ }\ \text{1}\text{.18}\ \text{g}\ \text{c}{{\text{m}}^{\text{3}}}\ \text{ }\!\!\times\!\!\text{ }\ \text{36%}\ \text{=}\ \text{0}\text{.4248}\ \text{g}\)
\(\displaystyle{\text{1}\text{.0}\ \text{c}{{\text{m}}^{\text{3}}}\ \text{原溶液中含溶質的摩爾數}\ \text{=}\ \frac{\text{0}\text{.4248}\ \text{g}}{\text{36}\text{.5}\ \text{g}\ \text{mo}{{\text{l}}^{-1}}}\ \text{=}\ \text{0}\text{.116}\ \text{mol}}\)
\(\displaystyle{\text{稀釋後溶液的摩爾濃度}\ \text{=}\ \frac{\text{0}\text{.116}\ \text{mol}}{\text{500}\text{.0}\ \text{c}{{\text{m}}^{\text{3}}}}\ \text{=}\ \frac{\text{0}\text{.116}\ \text{mol}}{\text{0}\text{.500}\ \text{d}{{\text{m}}^{\text{3}}}}\ \text{=}\ \text{0}\text{.023}\ \text{mol}\ \text{d}{{\text{m}}^{-3}}}\)
所以，稀釋後氫氯酸溶液的摩爾濃度為 \(\text{0}\text{.023}\ \text{mol}\ \text{d}{{\text{m}}^{-3}}\)。
實驗室中有已製備好的 \(\text{5}\text{.0}\ \text{M}\) 的氯化鋇標準溶液。現一實驗中需要製備 \(\text{500}\text{.0}\ \text{c}{{\text{m}}^{\text{3}}}\) 濃度為 \(\text{0}\text{.10}\ \text{M}\) 的氯化鋇溶液，計算所需量取原氯化鋇標準溶液的體積？(答案保留至小數點後第一位。）
與練習三類似，稀釋前後溶質的質量或摩爾數相等。但要先求出稀釋後溶液中溶質的摩爾數，再計算要含有這些溶質需要多少體積的濃溶液。
\(\begin{align*} \text{所需溶液中含有溶質的摩爾數}\ &= \ 500.0\ \text{c}{{\text{m}}^{3}}\ \times \ 0.10\ \text{M}\ \\ &= \ 0.500\ \text{d}{{\text{m}}^{3}}\ \times \ 0.10\ \text{mol}\ \text{d}{{\text{m}}^{-3}}\ \text{=}\ 0.050\ \text{mol} \end{align*}\)
\(\displaystyle{\text{所需的原濃溶液體積}\ \text{=}\ \frac{\text{0}\text{.050}\ \text{mol}}{\text{5}\text{.0}\ \text{M}}\ \text{=}\ \frac{\text{0}\text{.050}\ \text{mol}}{5.\text{0}\ \text{mol}\ \text{d}{{\text{m}}^{-3}}}\ \text{=}\ \text{0}\text{.01}\ \text{d}{{\text{m}}^{\text{3}}}\ \text{=}\ \text{10}\text{.0}\ \text{mL}}\)
所以，需要量取 \(\text{10}\text{.0}\ \text{mL}\) 的原氯化鋇溶液。