The p-Block Elements JEE 2019: Sulphur Question Solved

What is the answer to the p-Block Elements JEE 2019 sulphur question?

The required mass is 288.00 g in this p-block elements JEE 2019 question: one mole of rhombic sulphur contains eight moles of sulphur atoms, not one. The source is JEE Advanced 2019, Paper 2, Chemistry. This is a numerical-answer question, not multiple-choice.

The amount of water produced (in g) in the oxidation of one mole of rhombic sulphur by conc. nitric acid to a compound with the highest oxidation state of sulphur is ______.

The supplied molar mass of water is:

M (H_{2} O) = 18 g m o l^{- 1}

We need the mass of water formed when one mole of rhombic sulphur reacts with concentrated nitric acid and sulphur reaches its maximum oxidation state.

Why must we start with sulphur molecules rather than sulphur atoms?

Rhombic sulphur consists of eight-atom molecules, so the given amount is one mole of molecules containing eight moles of sulphur atoms. The word “mole” counts entities; identifying those entities fixes the starting amount.

1 m o l S_{8} contains 8 m o l o f s u l p h u r a t o m s

Elemental sulphur starts at oxidation state zero. Its highest oxidation state is positive six, and the sulphur-containing product in the official solution is sulphuric acid.

Check its oxidation state using hydrogen as positive one and oxygen as negative two:

H_{2} S O_{4} : 2 (+ 1) + x + 4 (- 2) = 0

$x = + 6$ Retain nitrogen dioxide as the reduction product in the supplied concentrated-acid reaction. The unbalanced framework is:

S_{8} + H N O_{3} \to H_{2} S O_{4} + N O_{2} + H_{2} O

How do we derive every coefficient in the balanced reaction?

One sulphur molecule requires 48 nitric acid molecules and produces 16 water molecules in the balanced equation. Derive these coefficients by setting the coefficient of rhombic sulphur to one, then balancing sulphur before the other elements.

Eight sulphur atoms require eight sulphuric acid molecules. Assign unknown coefficients to the remaining species:

S_{8} + a H N O_{3} \to 8 H_{2} S O_{4} + b N O_{2} + c H_{2} O

Nitrogen appears only in nitric acid and nitrogen dioxide. Each molecule contains one nitrogen atom, so: $b = a$

The sulphuric acid contributes 16 hydrogen atoms on the right. Water contributes twice its coefficient: $a = 16 + 2 c$

For oxygen, nitric acid contributes three atoms per molecule. On the right, sulphuric acid contributes 32 atoms, followed by oxygen in nitrogen dioxide and water: $3 a = 32 + 2 b + c$

Substitute the nitrogen balance:

3 a = 32 + 2 a + c \Rightarrow a = 32 + c

Both expressions for the nitric acid coefficient must agree:

16 + 2 c = 32 + c \Rightarrow c = 16

$a = b = 48$ The official balanced equation is:

S_{8} + 48 H N O_{3} \to 8 H_{2} S O_{4} + 48 N O_{2} + 16 H_{2} O

Check every element, not just sulphur. The left and right atom totals match:

S : & 8 = 8 H : & 48 = 16 + 32 = 48 N : & 48 = 48 O : & 144 = 32 + 96 + 16 = 144

How does the water coefficient give 288.00 g?

The balanced equation gives 16 moles of water per mole of rhombic sulphur. Multiply the given amount by that mole ratio, then convert water moles into grams. Keep the species beside each unit so the molecular basis stays explicit.

n (H_{2} O) = 1 m o l S_{8} \times \frac{16 m o l H_{2} O}{1 m o l S_{8}} = 16 m o l H_{2} O

Apply the mass relation: $m = n M$

m (H_{2} O) = 16 m o l \times 18 g m o l^{- 1} = 288 g

Required mass: 288.00 g. Numerical answer: 288.00.

The question asks for water generated by the reaction, not water initially present in the concentrated nitric acid solution. No solution-composition calculation is needed.

Why is 36 g the wrong answer?

The illustrative incorrect result of 36 g comes from changing the mole basis halfway through. It is not an official option: this numerical question contains no answer choices.

Dividing the official equation by eight gives an atom-basis stoichiometric shorthand:

S + 6 H N O_{3} \to H_{2} S O_{4} + 6 N O_{2} + 2 H_{2} O

The single sulphur symbol counts sulphur atoms. It does not claim that rhombic sulphur is monatomic.

The incorrect calculation is:

m_{incorrect} = 1 \times 2 \times 18 = 36 g

This applies a per-mole-of-sulphur-atoms ratio to one mole of sulphur molecules. Repair it by restoring the eight atoms per molecule:

1 m o l S_{8} = 8 m o l o f s u l p h u r a t o m s

m (H_{2} O) = 8 \times 2 \times 18 = 288 g

Retain the molecular equation throughout to avoid this conversion. Before using any coefficient, label the starting amount explicitly as moles of the molecular species:

n (S_{8}) = 1 m o l

Which three practice questions check the same method?

These questions check the starting entity, coefficient ratio and oxidation-state calculation separately. All three are original practice questions based on the chapter, not additional JEE 2019 questions or verified PYQs. Attempt each before reading its worked check.

Under the same reaction conditions, what mass of water forms from 0.25 mol of rhombic sulphur?

Keep the same products and balanced coefficients. Only the starting amount changes, so scale the water amount directly.

n (S_{8}) = 0.25 m o l

n (H_{2} O) = 0.25 \times 16 = 4 m o l

m (H_{2} O) = 4 \times 18 = 72 g

How many moles of concentrated nitric acid are consumed per mole of rhombic sulphur?

Read the reactant coefficient ratio directly. No molar mass is needed because the requested quantity is in moles.

\frac{n (H N O_{3})}{n (S_{8})} = \frac{48}{1}

The answer is 48 moles of nitric acid per mole of rhombic sulphur.

What is sulphur’s oxidation state in sulphur dioxide, sulphur trioxide and sulphuric acid?

Set the sum of oxidation states to zero for each neutral compound.

S O_{2} : x - 4 = 0 \Rightarrow x = + 4

S O_{3} : x - 6 = 0 \Rightarrow x = + 6

H_{2} S O_{4} : 2 + x - 8 = 0 \Rightarrow x = + 6

The last two compounds both contain sulphur at its maximum oxidation state. Oxidation state alone does not identify the product: use the specified reaction, which gives sulphuric acid here.

Next, practise another same-chapter reaction in P-Block Elements JEE 2019: Gold Extraction Explained.

Frequently asked questions

What is the answer to the JEE Advanced 2019 rhombic sulphur question?

The required mass of water is 288.00 g. One mole of rhombic sulphur, S8, produces 16 moles of water on oxidation by concentrated nitric acid under the specified conditions. Using the supplied molar mass of 18 g/mol gives 16 × 18 = 288 g.

What is the balanced equation for rhombic sulphur with concentrated nitric acid?

The balanced equation is S8 + 48HNO3 → 8H2SO4 + 48NO2 + 16H2O. Sulphur goes from oxidation state 0 to +6 in sulphuric acid, while nitric acid is reduced to nitrogen dioxide.

Why is 36 g wrong in the JEE 2019 sulphur question?

The result 36 g uses the water ratio for one mole of sulphur atoms, not one mole of S8 molecules. One mole of rhombic sulphur contains eight moles of sulphur atoms, so the correct calculation is 8 × 2 × 18 = 288 g. This is a numerical-answer question; 36 g is an illustrative error, not an official option.

How many moles of nitric acid react with one mole of rhombic sulphur?

Under the specified concentrated nitric acid reaction conditions, one mole of S8 consumes 48 moles of HNO3. This follows directly from the 1:48 reactant coefficient ratio in the balanced equation.