Chemistry 5070 · O Level · Preparation of salts

Preparation of salts — practice question

This question concerns several sulfur compounds.
(a(i))[3]

A student titrates 25.0 cm$^3$ of dilute sulfuric acid using sodium hydroxide of concentration 0.0150 mol/dm$^3$, with litmus as the indicator. Exactly 24.0 cm$^3$ of aqueous sodium hydroxide neutralises the dilute sulfuric acid. Calculate the concentration of the dilute sulfuric acid.

(a(ii))[3]

Describe a method for obtaining pure, dry crystals of sodium sulfate from aqueous sodium sulfate.

(b)[1]

Concentrated sulfuric acid oxidises arsenic to arsenic(III) oxide. Complete the equation for this reaction. $\ldots\text{As} + \ldots\text{H}_2\text{SO}_4 \rightarrow \text{As}_4\text{O}_6 + \ldots\text{H}_2\text{O} + 6\text{SO}_2$

(c)[1]

Sulfur dichloride, Cl-S-Cl, is a simple molecular substance. Finish the dot-and-cross diagram for one molecule of sulfur dichloride. Show only the outer shell electrons.

(d)[2]

The melting point of sulfur dichloride is $-121\,^\circ\text{C}$. The boiling point of sulfur dichloride is $59\,^\circ\text{C}$. Deduce what state sulfur dichloride is in at room temperature. Give a reason for your answer.

Worked solution & mark scheme

This 10-mark question has a full step-by-step worked solution and mark scheme. One marking point: moles of NaOH = 0.0150 \times 24.0 / 1000 = 3.60 \times 10⁻⁴

  • Full mark scheme, point by point
  • Step-by-step worked solution
  • Write your answer & get it marked instantly by AI