Moles, molar mass and the Avogadro constant IB DP Physics Study Notes - 2025 Syllabus
Moles, molar mass and the Avogadro constant IB DP Physics Study Notes
Moles, molar mass and the Avogadro constant IB DP Physics Study Notes at IITian Academy focus on specific topic and type of questions asked in actual exam. Study Notes focus on IB Physics syllabus with Students should understand
- pressure as given by $P=\frac{F}{A}$ where $F$ is the force exerted perpendicular to the surface
- the amount of substance $n$ as given by $n=\frac{N}{N_{\mathrm{A}}}$ where $N$ is the number of molecules and $N_{\mathrm{A}}$ is the Avogadro constant
Standard level and higher level: 6 hours
Additional higher level: There is no additional higher level content
- IB DP Physics 2025 SL- IB Style Practice Questions with Answer-Topic Wise-Paper 1
- IB DP Physics 2025 HL- IB Style Practice Questions with Answer-Topic Wise-Paper 1
- IB DP Physics 2025 SL- IB Style Practice Questions with Answer-Topic Wise-Paper 2
- IB DP Physics 2025 HL- IB Style Practice Questions with Answer-Topic Wise-Paper 2
Pressure
- Consider a gas molecule confined in a closed box as shown.
- Since the molecule’s direction is always changing, it is accelerating, and thus feeling a force from the walls of the box.
- From Newton’s third law, the molecule is exerting an equal and opposite force on the walls of the box.
- The walls of the box feel a pressure p, given by the formula below:
- From the formula we can see that the units of pressure are (N m-2), also known as Pascals (Pa).
PRACTICE:
A 150-kg man stands on one foot on some ice. Given that his foot is about 9.0 cm by $10 . \mathrm{cm}$ in rectangular cross-section, find the pressure on the ice.
▶️Answer/Explanation
SOLUTION:
First, find the force:
$$F=m g=150(9.8)=1470 \mathrm{~N}$$
Then find the area:
$$A=L W=(.09)(.10)=0.009 \mathrm{~m}^2$$
Finally
$$p=\frac{F}{A}=\frac{1470}{0.009}=160000 \mathrm{Nm}^{-2}=160000 \mathrm{~Pa} .$$
The mole and molar mass
- Recall from the periodic table of the elements that each element has certain numbers associated with it.
- We define the mole of a homogeneous substance as follows:
EXAMPLE:
Find the mass (in kg) of one mole of carbon.
▶️Answer/Explanation
SOLUTION:
From the periodic table we see that it is just 1 mole C = 12.011 grams = 0.012011 kg.
PRACTICE:
Water is made up of 2 hydrogen atoms and 1 oxygen atom and has a molecular formula given by $\mathrm{H}_2 \mathrm{O}$. Find
(a) the gram atomic weight of water.
(b) the mass in grams of 1 mole of water.
(c) how many moles of hydrogen and oxygen there are in 1 mole of water.
SOLUTION:
(a) The GAW of $\mathrm{H}_2 \mathrm{O}$ is given by
$$
\begin{aligned}
& 2(1.00794)+1(15.9994)=18.01528 \mathrm{~g} \text { per mole. }
\end{aligned}
$$
(b) Thus the mass of 1 mole of $\mathrm{H}_2 \mathrm{O}$ is 18.01528 g .
(c) Since each mole of $\mathrm{H}_2 \mathrm{O}$ has 2 H and 1 O , there are 2 moles of H and 1 mole of O for each mole of water.
The Avogadro constant
- It turns out the a mole of carbon (12 g) and a mole of phosphorus (31 g) has the same number of atoms.
- This means that 30.973762 g of P has the same number of atoms and 12.011 g of C.
- That number, NA , is given here:
- To find the number of atoms in a sample, simply convert sample to moles, then use the conversion 1 mol = 6.02×1023 molecules.
- To find the number of moles n in a sample containing N atoms, use this formula:
EXAMPLE:
How many atoms of P are there in 31.0 g of it? How many atoms of C are there in 12.0 g of it?
▶️Answer/Explanation
SOLUTION:
There are NA = 6.02×1023 atoms of P in 31.0 g of it.
There are NA = 6.02×1023 atoms of C in 12.0 g of it.
EXAMPLE:
A sample of carbon has $1.28 \times 10^{24}$ atoms as counted by Marvin the Paranoid Android.
a) How many moles is this?
b) What is its mass?
▶️Answer/Explanation
SOLUTION:
It is best to start with the given quantity.
a) $\left(1.28 \times 10^{24}\right. atoms )\left(\frac{1 \mathrm{~mol}}{6.02 \times 10^{23} \text { atoms }}\right)=2.13 \mathrm{~mol}$.
b) $\quad(2.13 ~~\mathrm{mol})\left(\frac{12.011 \mathrm{~g}}{mol}\right)=25.5 \mathrm{~g ~of~ c}$
EXAMPLE:
A sample of carbon has $1.28 \times 10^{24}$ atoms as counted by Marvin the Paranoid Android.
a) How many moles is this?
b) What is its mass?
▶️Answer/Explanation
SOLUTION:
It is best to start with the given quantity.
a) $\left(1.28 \times 10^{24}\right. atoms )\left(\frac{1 \mathrm{~mol}}{6.02 \times 10^{23} \text { atoms }}\right)=2.13 \mathrm{~mol}$.
b) $\quad(2.13 ~~\mathrm{mol})\left(\frac{12.011 \mathrm{~g}}{mol}\right)=25.5 \mathrm{~g ~of~ c}$