Chemistry 2012 Jamb Examination

The number of atoms can be determined using the relationship between mass, molar mass, and Avogadro’s constant.

First, calculate the number of moles of magnesium:\[ n = \dfrac{\text{mass}}{\text{molar mass}} = \dfrac{6.0}{24} = 0.25 \; \text{mol} \]
The number of atoms is then obtained by multiplying the number of moles by Avogadro’s number:\[ \text{Number of atoms} = n \times N_A \]\[ \text{Number of atoms} = 0.25 \times 6.02 \times 10^{23} \]\[ = 1.505 \times 10^{23} \; \text{atoms} \]
This means there are approximately \(1.51 \times 10^{23}\) atoms of magnesium present in 6.0 g of the element.

When gas is collected over water, the total pressure includes the vapour pressure of water. To find the pressure of the dry gas:\[ P_1 = 765 - 5 = 760 \; \text{mmHg} \]
We apply Boyle’s law and Charles’ law combined (the gas laws):\[ \dfrac{P_1 V_1}{T_1} = \dfrac{P_2 V_2}{T_2} \]Given:
\( P_1 = 760 \; \text{mmHg} \), \( V_1 = 50 \; \text{cm}^3 \), \( T_1 = 10 + 273 = 283 \; \text{K} \)
At s.t.p., \( P_2 = 760 \; \text{mmHg} \) and \( T_2 = 273 \; \text{K} \)

Substitute:\[ \dfrac{760 \times 50}{283} = \dfrac{760 \times V_2}{273} \]\[ V_2 = 50 \times \dfrac{273}{283} \]\[ V_2 = 48.23 \; \text{cm}^3 \]
The gas therefore occupies 48.23 cm³ at standard temperature and pressure.

At constant temperature, the relationship between pressure and volume of a gas is defined by Boyle’s law, which states that:\[ P \propto \dfrac{1}{V} \; \text{or} \; P_1V_1 = P_2V_2 \]
This means that as the pressure increases, the volume decreases proportionally, provided the temperature remains constant. The gas molecules are forced closer together, reducing the space they occupy. Although the number of effective collisions per unit area increases, the overall volume of the gas decreases. Hence, an increase in pressure leads to a corresponding decrease in volume.