(a)(i) State Graham's law of diffusion (ii) If \(100\ \text{cm}^3\) of oxygen diffused in 4 seconds and \(50\text{cm}^3\) of gas Y diffused in 3 seconds, ca...

Assessment: WAEC SSCE - Chemistry - 2005 (Essay) Subject: Chemistry

Question 1 Report

(a)(i) State Graham's law of diffusion

(ii) If \(100\ \text{cm}^3\) of oxygen diffused in 4 seconds and \(50\text{cm}^3\) of gas Y diffused in 3 seconds, calculate the relative molecular mass of gas Y. (0 = 16)

(b) Consider the following equilibrium reaction: \(X + 2Y_{(g)} \rightleftharpoons XY_{2(9)}\) \(\Delta H = -52\text{KJ mol}^{-1}\)

(i) State what happens to the yield of \(XY_2\) when the temperature is increased

(ii) Explain the effect of decrease in pressure on the equilibrium position.

(iii) State the effect of a catalyst on the I. position of equilibrium II. activation energy

(c)(i) State the differences between the solubilities of solids and gases in liquids.

(ii) Name the physical-properties used it choosing separation techniques for the following mixtures:

I. kerosene and petrol II. calcium trioxocarbonate (IV) and potassium chloride. III. ammonium chloride and sodium chloride.

(d)(i) State a method of preparing each of the following salts:

Acid Basicity
\(H_3PO_4\)
\(CH_3COOH\)
\(HNO_2\)

(iii) State the difference between anhydrous and hydrated salts.

Answer Details

(a)(i) Graham's law of diffusion

At constant temperature and pressure, the rate of diffusion of a gas is inversely proportional to the square root of its density (or relative molecular mass).

\[ \text{rate} \propto \frac{1}{\sqrt{M}} \]

(a)(ii) Relative molecular mass of gas Y

Rate of oxygen \( = \dfrac{100}{4} = 25\ \text{cm}^3\text{s}^{-1} \); rate of Y \( = \dfrac{50}{3} = 16.67\ \text{cm}^3\text{s}^{-1} \). Relative molecular mass of \( O_2 = 2 \times 16 = 32 \).

\[ \frac{R_{O_2}}{R_Y} = \sqrt{\frac{M_Y}{M_{O_2}}} \]\[ \frac{25}{16.67} = \sqrt{\frac{M_Y}{32}} \;\Rightarrow\; 1.5 = \sqrt{\frac{M_Y}{32}} \]

Squaring both sides: \( 2.25 = \dfrac{M_Y}{32} \), so \( M_Y = 2.25 \times 32 = 72 \).

Relative molecular mass of gas Y = 72.

(b) \( X + 2Y_{(g)} \rightleftharpoons XY_{2(g)} \), \( \Delta H = -52\ \text{kJ mol}^{-1} \) (the forward reaction is exothermic).

(i) An increase in temperature favours the endothermic (backward) direction, so the yield of \( XY_2 \) decreases.

(ii) A decrease in pressure shifts the equilibrium towards the side with the greater number of gas molecules. The left side has 3 gas molecules \( (X + 2Y) \) while the right has 1 \( (XY_2) \); the position of equilibrium therefore moves backward (to the left) and the yield of \( XY_2 \) decreases.

(iii) A catalyst: I. has no effect on the position of equilibrium; II. lowers the activation energy of both the forward and backward reactions equally.

(c)(i) Solubility of solids compared with gases in liquids

  • The solubility of most solids increases with a rise in temperature, whereas the solubility of gases decreases with a rise in temperature.
  • Pressure has almost no effect on the solubility of solids, but the solubility of gases increases markedly with an increase in pressure.

(c)(ii) Physical property used for separation

  • I. Kerosene and petrol: difference in boiling points (fractional distillation).
  • II. Calcium trioxocarbonate(IV) and potassium chloride: difference in solubility in water (\( CaCO_3 \) insoluble, \( KCl \) soluble); dissolve, then filter.
  • III. Ammonium chloride and sodium chloride: difference in the ability to sublime (\( NH_4Cl \) sublimes); sublimation.

(d)(ii) Basicity of the acids

Basicity is the number of replaceable hydrogen ions produced per molecule of the acid on ionization.

AcidBasicity
\( H_3PO_4 \)3
\( CH_3COOH \)1
\( HNO_2 \)1

Only the hydrogen of the \( -COOH \) group in ethanoic acid is replaceable, so its basicity is 1.

(d)(iii) Anhydrous and hydrated salts

An anhydrous salt contains no water of crystallization, while a hydrated salt contains a fixed number of molecules of water of crystallization chemically combined with it (for example \( CuSO_4 \) is anhydrous while \( CuSO_4 \cdot 5H_2O \) is hydrated).

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