# Molar Volume & Molar Concentration

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Molar Volume is the volume occupied by one mole of gas.

• Molar volume changes with temperature and pressure. (Higher temperature = higher volume, higher pressure = smaller volume)
• Molar volume at a fixed temperature and pressure is the same for ALL gases.
• At room temperature ($25^{\circ}\text{C}$) and pressure (1 atmosphere, atm), also known as r.t.p., the molar volume of ANY gas is 24 $\text{dm}^{3}$.
• At standard temperature ($0^{\circ}\text{C}$) and pressure (1 atm), also known as s.t.p., the molar volume of ANY gas is 22.4 $\text{dm}^{3}$.
• $\text{dm}^{3}$ is known as cubic decimetre. 1 $\text{dm}^{3}$ = 1 litre = 1000 $\text{cm}^{3}$ = $\frac{1}{1000} \, \text{m}^{3}$

Note 1: It is important to know how to convert among $\text{cm}^{3}$, $\text{dm}^{3}$ and $\text{m}^{3}$ A short tutorial on the conversion process will be given at the end of this post. (For those who do not want to memorise the conversion numbers above)

Note 2: It is important to take note of the temperature and pressure in the given question to determine if r.t.p or s.t.p should be used.

At r.t.p,

$$\text{Vol. of gas} = \text{No. of moles} \times 24 \, \text{dm}^{3}$$

At s.t.p,

$$\text{Vol. of gas} = \text{No. of moles} \times 22.4 \, \text{dm}^{3}$$

Avogadro’s law states that, at the SAME temperature and pressure, equal volumes of all gases contain the same number of gas particles.

• $24 \, \text{dm}^{3}$ of helium gas at r.t.p. contains 1 mole of He atoms ($6 \times 10^{23}$ atoms)
• $24 \, \text{dm}^{3}$ of carbon dioxide gas at r.t.p. contains 1 mole of $\text{CO}_{2}$ atoms ($6 \times 10^{23}$ atoms)

### Example

Calculate the number of moles of gas particles for $200 \, \text{cm}^{3}$ of carbon dioxide gas at r.t.p.

• Convert to $\text{dm}^{3}$: \begin{aligned} 200 \, \text{cm}^{3} &= \frac{200}{1000} \, \text{dm}^{3} \\ &= 0.20 \, \text{dm}^{3} \end{aligned}
• Number of moles: $\frac{0.20}{24} = 0.0083 \, \text{mol}$

Calculate the number of gas particles for $10 \, \text{dm}^{3}$ of hydrogen gas at s.t.p.

• No need to convert since it is already in $\text{dm}^{3}$
• Number of moles: $\frac{10}{22.4} = 0.446 \, \text{mol}$
• Number of gas particles: $0.446 \times \left( 6 \times 10^{23} \right) = 2.67 \times 10^{23}$

### Molar Concentration

Molar concentration is the amount of solute (Typically in grams or moles) present in a given volume of solution. (Typically, $1 \, \text{dm}^{3}$ in Chemistry)

Hence,

$$\text{Conc.} \,(\text{g dm}^{-3}) = \frac{\text{Mass of solute in grams}}{\text{Vol. of solution in dm}^{3}}$$

$$\text{Conc.} \,(\text{mol dm}^{-3}) = \frac{\text{No. of moles of solute}}{\text{Vol. of solution in dm}^{3}}$$

Note: Concentration in $\text{mol dm}^{-3}$ is called molarity.

A molar solution refers to a solution that contains 1 mole of solute in 1 $\text{dm}^{3}$ of solution.

Important: If a solution with X $\text{mol dm}^{-3}$ is diluted with more solution, the number of moles of solutes does not change! The concentration will drop but the number of moles of solutes does not change. You can find the number of moles of the solutes and divide by the final volume of the solution to find the final concentration.

### Example

10.4 g of potassium hydroxide was dissolved to formed $250 \, \text{cm}^{3}$ of solution. Calculate the concentration of solution in $\text{g dm}^{-3}$ and $\text{mol dm}^{-3}$

• Convert to $\text{dm}^{3}$: $250 \, \text{cm}^{3} = \frac{250}{1000} = 0.250 \, \text{dm}^{3}$
• Conc. in $\text{g dm}^{-3}$: $\frac{10.4}{0.250} = 41.6 \, \text{g dm}^{-3}$
• Molar mass of potassium hydroxide: $39 + 16 + 1 = 56 \, \text{g mol}^{-1}$
• No. of moles of potassium hydroxide present: $\frac{10.4}{56} = 0.186 \, \text{mol}$
• Molarity: $\frac{0.186}{0.250} = 0.743 \, \text{mol dm}^{-3}$

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### How To Convert Among $\text{dm}^{3}$, $\text{cm}^{3}$ and $\text{m}^{3}$

In order to know how to convert among those units, you will need to learn what is prefixes. (Visit that link to learn more)

From prefixes, we know that: $1 \, \text{m} = 10 \, \text{dm} = 100 \, \text{cm}$

Recall that volume is $\text{length} \times \text{length} \times \text{length}$.

$1 \, \text{m} \times 1 \, \text{m} \times 1 \, \text{m} = 1 \, \text{m}^{3}$

$10 \, \text{dm} \times 10 \, \text{dm} \times 10 \, \text{dm} = 1000 \, \text{dm}^{3}$

$100 \, \text{cm} \times 100 \, \text{cm} \times 100 \, \text{cm} = 1000000 \, \text{cm}^{3}$

Hence, $1 \, \text{m}^{3} = 1000 \, \text{dm}^{3} = 1000000 \, \text{cm}^{3}$ 