物质的量浓度Molar concentration

注意 V 是溶液的体积而不是溶剂的体积，单位要换成 dm³：25.0 cm³ = 0.0250 dm³。例：把 4.00 g NaOH（M = 40.0 g mol⁻¹）溶解配成 250 cm³ 溶液，n = 0.100 mol，c = 0.100 ÷ 0.250 = 0.400 mol dm⁻³。质量浓度（g dm⁻³）= 物质的量浓度 × 摩尔质量。

V is the volume of the solution, not the solvent, and must be in dm³: 25.0 cm³ = 0.0250 dm³. Example: 4.00 g of NaOH (M = 40.0 g mol⁻¹) made up to 250 cm³ gives n = 0.100 mol and c = 0.100 ÷ 0.250 = 0.400 mol dm⁻³. Mass concentration (g dm⁻³) = molar concentration × molar mass.

准确称取溶质，在烧杯中用少量蒸馏水溶解，转移到容量瓶中，用蒸馏水洗涤烧杯和玻璃棒并把洗液一并转入，加水至接近刻度线后改用胶头滴管定容，使凹液面最低点与刻度线相切，最后塞紧瓶塞反复颠倒摇匀。

Weigh the solute accurately, dissolve it in a little distilled water in a beaker, and transfer it to a volumetric flask, rinsing the beaker and rod into the flask. Add water to near the mark, then use a dropper until the bottom of the meniscus sits on the line, stopper the flask and invert it repeatedly to mix.

稀释前后溶质的物质的量不变：c₁V₁ = c₂V₂。例如把 25.0 cm³ 2.00 mol dm⁻³ 的溶液稀释到 250 cm³，浓度变为 0.200 mol dm⁻³。由于溶液体积随温度变化，物质的量浓度也随温度略有变化；需要与温度无关的浓度时可改用质量摩尔浓度（mol kg⁻¹）。

The amount of solute is unchanged on dilution: c₁V₁ = c₂V₂. Diluting 25.0 cm³ of a 2.00 mol dm⁻³ solution to 250 cm³ gives 0.200 mol dm⁻³. Because solution volume changes with temperature, molar concentration does too; where a temperature-independent measure is needed, molality (mol kg⁻¹) is used instead.