NCERT Solutions for Class 9th Science Chapter 9 Project work — The Journey Beyond

Book page 183 Updated on2026-09-08

Q1.

Design and perform an experiment to show and compare that water always contains hydrogen and oxygen in the same ratio, regardless of its source.

Answer

Aim: to show that water from different sources always splits into hydrogen and oxygen in the same fixed ratio — a direct test of the Law of Constant Proportions.

Principle. Passing a direct current through water decomposes it. Hydrogen collects at the cathode (negative electrode) and oxygen at the anode. If water is always H₂O, the volume ratio must always be 2 : 1, whatever the source.

Apparatus: a Hofmann voltameter (or a plastic tub with two inverted graduated test tubes over two carbon electrodes), a 9 V battery, dilute sulfuric acid, and water samples from four sources — tap, rain, borewell and river.

Method

Filter and then distil each sample. This is the step that decides the experiment — the dissolved salts must go, or you will be electrolysing a solution, not water.
Fill the apparatus with the first distilled sample. Add 3–4 drops of dilute sulfuric acid to make it conduct. Add exactly the same amount to every sample.
Fill both graduated tubes completely, invert them over the electrodes, and connect the 9 V battery.
Run for a fixed time — say 10 minutes — and read the volume of gas collected in each tube.
Test the gases: the larger volume gives a squeaky pop with a lighted splint (hydrogen); the smaller one relights a glowing splint (oxygen).
Repeat, washing the apparatus thoroughly, for each of the four sources.

Sample observations

Source of water	Volume of H₂	Volume of O₂	Ratio H₂ : O₂
Tap water	20.0 mL	10.0 mL	2 : 1
Rain water	18.2 mL	9.1 mL	2 : 1
Borewell water	21.4 mL	10.6 mL	2 : 1
River water	19.6 mL	9.9 mL	2 : 1

Conclusion. The total volume varies from run to run — that only reflects small differences in current and time — but the ratio is 2 : 1 every single time. Water from every source has the same composition.

2 volumes of H₂ : 1 volume of O₂
→ 2 molecules of H₂ : 1 molecule of O₂ → 4 H atoms : 2 O atoms → H₂O
By mass: (4 × 1 u) : (2 × 16 u) = 4 : 32 = 1 : 8 ✓

Controls to keep the test honest: the same volume of acid in every run, the same electrodes, the same current and the same time; and read the volumes only after the tubes have cooled and the bubbles have cleared.

Safety first: hydrogen is flammable — keep flames away except for the deliberate splint test, and do the test on a small sample only. Use a low-voltage battery and never a mains supply.

Q2.

Compare atoms and ions of any three elements. Show the number of electrons before and after ion formation using bar graphs.

Answer

Take sodium, magnesium and chlorine — two metals that lose electrons and one non-metal that gains one.

Element	Z (protons)	Atom: electrons and configuration	Change	Ion: electrons and configuration	Charge
Sodium	11	11 — 2, 8, 1	loses 1 e⁻	10 — 2, 8	Na⁺
Magnesium	12	12 — 2, 8, 2	loses 2 e⁻	10 — 2, 8	Mg²⁺
Chlorine	17	17 — 2, 8, 7	gains 1 e⁻	18 — 2, 8, 8	Cl⁻

Electrons before and after ion formation. The bar falls for the two metals and rises for chlorine — but every ion ends on a full octet.

What the graph shows. Sodium and magnesium drop to 10 electrons; chlorine climbs to 18. The numbers move in opposite directions, yet all three land on the same kind of arrangement — a complete outermost shell. Na⁺ and Mg²⁺ both become 2, 8 (like neon) and Cl⁻ becomes 2, 8, 8 (like argon).

Why it happens: the number of protons never changes — sodium still has 11, magnesium 12, chlorine 17. Only the electron count moves, and the imbalance that this creates is the charge on the ion. This is why an ion behaves so differently from its parent atom: sodium metal catches fire in water, while Na⁺ is what you sprinkle on your food.

Try This: extend the graph with a third bar for protons in each case. It stays flat at 11, 12 and 17 while the electron bars move — a picture of exactly why the charge appears.

Q3.

Make a card game with cations and anions. Possible ideas may include — Pick cards from the pile or from open cards and match them with cards in your hand to form compounds. You may discard any unwanted cards. Ask other players for cards and form compounds using the cards they already have.

Answer

Here is a complete, playable design you can adapt.

The deck (40 cards). Print each ion on a card with its symbol, charge and a colour code — green for cations, red for anions.

Type	Ions	Copies of each
Cations, 1+	Na⁺, K⁺, NH₄⁺	3 each
Cations, 2+	Ca²⁺, Mg²⁺, Cu²⁺	2 each
Cations, 3+	Al³⁺, Fe³⁺	2 each
Anions, 1−	Cl⁻, OH⁻, NO₃⁻	3 each
Anions, 2−	O²⁻, SO₄²⁻, CO₃²⁻	2 each
Anions, 3−	PO₄³⁻	2

Rules

Deal 7 cards to each of 2–4 players. Place the rest face down as the pile and turn one card face up beside it.
On your turn, either draw the top card of the pile or take the face-up card. Then discard one card face up.
At any time you may lay down a set of cards that forms a real compound — but only if the total positive charge exactly cancels the total negative charge. Say the compound's name and write its formula.
You may also ask one other player for a specific ion (“Do you have a sulfate?”). If they have it they must hand it over, and they then draw a replacement from the pile.
The round ends when a player has no cards left.

Scoring — the part that teaches. Score the number of cards used in the compound, so bigger, harder combinations pay more.

Cards laid down	Formula	Charge check	Points
Na⁺ + Cl⁻	NaCl	(+1) + (−1) = 0	2
Ca²⁺ + 2 Cl⁻	CaCl₂	(+2) + 2(−1) = 0	3
2 Al³⁺ + 3 O²⁻	Al₂O₃	2(+3) + 3(−2) = 0	5
3 Ca²⁺ + 2 PO₄³⁻	Ca₃(PO₄)₂	3(+2) + 2(−3) = 0	5

Challenge rule: any player may challenge a formula. If the charges do not cancel, or the brackets are wrong, the set is returned to the owner's hand and the challenger scores 2 points. This one rule is what forces everybody to check the arithmetic.

Why it works as a learning game: you cannot lay a single valid set without doing the criss-cross in your head. Playing a few rounds drills charge balancing, bracket use and ion names far more effectively than copying formulae from a list.

Q4.

To learn more about molecules you can explore the link given below —

Answer

The link is to PhET's Build a Molecule simulation from the University of Colorado Boulder:

https://phet.colorado.edu/sims/html/build-a-molecule/latest/build-a-molecule_all.html

What it lets you do. You drag atoms — H, O, C, N, Cl and others — out of buckets and snap them together. The simulation accepts a join only if it is chemically possible, so you cannot build H₃O or CO₃ by force. When a legal molecule is complete it names itself and shows you a 3-D model you can rotate.

Try these in order, and predict before you click.

Build	What to notice	Links back to
H₂, O₂, N₂	Two identical atoms are enough — a molecule of an element	Section 9.4.1 A
H₂O, then try H₃O	H₃O is refused. Oxygen accepts exactly two bonds	Law of Constant Proportions
CH₄, NH₃, H₂S	Carbon takes 4 bonds, nitrogen 3, sulfur 2 — the shortfall rule	Q10, Pause and Ponder p. 171
CO, then CO₂	Same two elements, two different fixed ratios	Section 9.4.1 C, naming
Rotate the 3-D model of H₂O	The molecule is bent, not straight — a fact the flat dot diagram hides	Fig. 9.10

One thing to be careful about. The simulation builds molecules, so it handles covalent compounds well. It will not build NaCl, because sodium chloride is not a molecule at all — it is a 3-D lattice of Na⁺ and Cl⁻ ions, and NaCl is only its formula unit. That limitation is itself worth noticing: it is section 9.4.2 in action.

Try This: keep a notebook as you play. For every molecule you succeed in building, write the formula and work out its molecular mass by hand, then check whether your total agrees with the sum of the atoms you dragged in.

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