Gas volumes and concentration
Introduced| English | Meaning |
|---|---|
| concentration/ˌkɒnsənˈtreɪʃn/ | The amount of a substance per unit volume of a mixture or solution. |
| solute/ˈsɒljuːt/ | A substance dissolved in a solvent to form a solution. |
| molar gas volume/ˈməʊlə ɡæs ˈvɒljuːm/ | The volume occupied by one mole of gas at a specified temperature and pressure. |
Does a gas syringe measure the same thing as a measuring flask?
- A gas volume can give the amount of gas; a solution volume needs a concentration 浓度 as well.
- Concentration means the amount of dissolved solute 溶质 in a given final solution volume, not the volume of solvent added.
Gas volumes at stated conditions
- Supplement: use the molar gas volume 摩尔气体体积 of $24\ \mathrm{dm^3}$ per mole at room temperature and pressure, r.t.p., for this course's calculations.
- Use $n=V/24$ with $V$ in dm³, or $n=V/24000$ with $V$ in cm³. At other conditions, use the molar gas volume supplied in the question.
At r.t.p., 2.0 mol of gas occupies $V=24n=24\times2.0=48\ \mathrm{dm^3}$. A separate 600 cm³ sample is 0.600 dm³, so $n=0.600/24=0.0250\ \mathrm{mol}$. Using 600 ÷ 24 would mix cm³ and dm³.
What volume (dm³) does 3 moles of a gas occupy at rtp (1 mole = 24 dm³)?
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volume = 3 × 24 = 72 dm³.
How many moles of gas are in 48 dm³ at rtp?
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moles = 48 ÷ 24 = 2.
At room temperature and pressure, one mole of any gas occupies ______ dm³.
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The molar gas volume at rtp is 24 dm³.
A gas occupies 360 cm³ at r.t.p. Use 24 dm³ per mol. Find its amount in mol.
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360 cm³ = 0.360 dm³. n=0.360/24=0.0150 mol.
Two ways to express concentration
- Core: concentration can be measured in $\dfrac{\mathrm{g}}{\mathrm{dm^3}}$ or $\dfrac{\mathrm{mol}}{\mathrm{dm^3}}$. Mass concentration is mass of solute ÷ final solution volume.
- Supplement: molar concentration is $c=n/V$. Rearrange to $n=cV$ or $V=n/c$, using $V$ in dm³. A volume of 250 cm³ is 0.250 dm³.
A solution contains 0.50 mol of solute in a final volume of 2.0 dm³. Find its concentration in $\dfrac{\mathrm{mol}}{\mathrm{dm^3}}$.
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concentration = 0.5 ÷ 2 = 0.25 $\dfrac{\mathrm{mol}}{\mathrm{dm^3}}$.
A solution contains 4.0 g solute in a FINAL volume of 200 cm³. Find its mass concentration in $\dfrac{\mathrm{g}}{\mathrm{dm^3}}$.
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200 cm³ = 0.200 dm³. Mass concentration = 4.0/0.200 = 20 $\dfrac{\mathrm{g}}{\mathrm{dm^3}}$.
Calculate from the final solution volume
- A solution contains 5.85 g of NaCl in a final volume of 0.500 dm³. Mass concentration $=5.85/0.500=11.7\ \dfrac{\mathrm{g}}{\mathrm{dm^3}}$.
- Its molar mass is $58.5\ \dfrac{\mathrm{g}}{\mathrm{mol}}$, so $n=5.85/58.5=0.100\ \mathrm{mol}$ and $c=0.100/0.500=0.200\ \dfrac{\mathrm{mol}}{\mathrm{dm^3}}$. The two concentrations describe the same solution with different units.
The 24 dm³ rule is for a gas at r.t.p., not a solution. A solution containing 1 mol of solute does not automatically occupy 24 dm³. Convert cm³ to dm³ before using concentration in moles or grams per dm³.
Before using the concentration formula, a volume in cm³ should be:
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1 dm³ = 1000 cm³.
Find the amount in 25.0 cm³ of a 0.200 $\dfrac{\mathrm{mol}}{\mathrm{dm^3}}$ solution, in mol.
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V=0.0250 dm³. n=cV=0.200×0.0250=0.00500 mol.
Explain dilution without losing solute
- If no solute is lost or reacts, adding water increases the final volume but leaves the number of moles unchanged. Concentration falls.
- Doubling the final volume halves the concentration. Use $n=cV$ to check the amount before and after; doubling the added water is not always doubling the final volume.
Water is added to 50.0 cm³ of 0.400 $\dfrac{\mathrm{mol}}{\mathrm{dm^3}}$ solution to give a final volume of 200 cm³. No solute is lost. Find the new concentration in $\dfrac{\mathrm{mol}}{\mathrm{dm^3}}$.
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Initial n=0.400×0.0500=0.0200 mol. Final c=0.0200/0.200=0.100 $\dfrac{\mathrm{mol}}{\mathrm{dm^3}}$.
Match the measurement to its method
- For gas, identify the conditions and convert the volume. For solution, identify whether the concentration uses grams or moles per dm³.
- State the final unit and check the size: a 25 cm³ portion is one fortieth of a dm³, so it contains one fortieth of the solute in a dm³ of the same solution.
Gas at r.t.p.: $n=V/24$ for dm³. Solution: $n=cV$. Mass concentration uses grams instead of moles. In each case, name the substance and use matching units.
A gas sample contains 5.0 mol at r.t.p. Use 24 dm³ per mol. Calculate its volume and explain why this rule cannot give the volume of a solution containing 5.0 mol solute.
Gas molar volume and solution concentration are different relationships.
A gas contains 5.0 mol at r.t.p. Using 24 dm³ per mol, what volume does it occupy?
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Five times 24 is 120.
A question gives a gas molar volume of 22.0 dm³ per mol at its stated conditions. Which value should you use?
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24 dm³ per mol is the course r.t.p. value. Use a supplied value for other conditions.