QCC Notes
CLASS 11 · CHEMISTRYJEE MAIN × NEETहिंदी
§ 1.4NCERT Class 11 · Chemistry · Chapter 1

Uncertainty in Measurement: Significant Figures Done Right

Every measurement carries an honest uncertainty — and significant figures are how chemistry reports it. Count them, round them, carry them through calculations, and never let an answer claim more precision than the data earned.

01

Why Measurements Are Uncertain — Complete Theory

No balance reads “exactly”. A digit like 2.50 g means the last zero is a judgement — the instrument’s edge of certainty. Significant figures encode this honestly: all reliably known digits plus one uncertain digit. Write 2.50 g and you claim ±0.01 g precision; write 2.500 g and you claim ten times better — same substance, different confession. This is why an answer can never be more precise than its least precise input.

Two vocab words that exams keep separating: precision — how close repeated measurements are to each other; accuracy — how close they are to the true value. A balance can be beautifully repeatable and consistently wrong (a calibration error makes it precise, not accurate). The target diagram below makes the distinction visual.

Scientific notationN × 10ⁿ, 1 ≤ N < 10 × ÷ ruleleast s.f. wins + − ruleleast decimals win

Scientific notation writes any number as N × 10ⁿ with 1 ≤ N < 10 — 0.0034 becomes 3.4 × 10⁻³. Its gift: every digit of N is significant, so notation declares precision unambiguously. That is the cure for the ambiguous-trailing-zero trap below. Multiplication adds exponents; division subtracts them.

Table 1 — Counting rules with graded examples
NumberSig figsRule applied
4253All non-zero digits count
20054Zeros between non-zeros count
0.00342Leading zeros never count
2.5004Trailing zeros after a decimal count
0.020504Leading skip; 2, 0, 5 and final 0 count
2500ambiguousWhole-number trailing zeros — write 2.5 × 10³ (2 s.f.) or 2.500 × 10³ (4)
2 apples; 100 cm = 1 munlimitedExact numbers (counts, defined factors) never limit precision

Calculation rules pick which input governs the answer. For multiplication and division: the result keeps the fewest significant figures — 4.2 × 1.40 = 5.88 → 5.9 (2 s.f., because 4.2 has two). For addition and subtraction: the result keeps the fewest decimal places — 3.9 + 4.52 = 8.42 → 8.4 (1 decimal, because 3.9 has one). Notice the asymmetry: ×÷ counts digits, +− counts places. Mixing these up is the most common sig-fig error in JEE numeric-answer questions.

Finally, dimensional analysis (the factor-label method) — the safest unit-conversion machinery ever invented. Multiply by conversion factors written as fractions equal to 1 (100 cm / 1 m), arranging them so unwanted units cancel diagonally. Convert 1 L to m³: 1 L × (10⁻³ m³ / 1 L) = 10⁻³ m³. If the units don’t cancel to what the question wants, the setup is wrong — the method self-reports its own errors, which is why § 1.10 chains lean on it heavily.

02

Visualising Precision & Counting Live

Chaar targets, ek scanner — the accuracy/precision grid first, then scan or round any number you like.

Accuracy vs precision — where the shots land decides which one you have FIG. 1 — ACCURACY × PRECISION, THE TARGET GRID Accurate + Precise calibrated instrument, careful hand Precise, not accurate tight cluster, wrong corner — systematic error Accurate on average mean hits the bullseye, spread is wild Precision = shots agree with each other. Accuracy = shots agree with the truth. the fourth quadrant (“neither”) needs no diagram — it needs a better lab
FIG. 1 — The classic four: both (calibrated + careful), precise-but-inaccurate (a systematic error the repeats can’t see), accurate-on-average (errors cancel only in the mean), and neither. Sig figs report the precision axis; calibration decides the accuracy axis.
Try it live

Sig-Fig Scanner

Verdict — scanner

Rounding result

Underlined digits = significant · grey underline = leading zeros (never count) · dashed = ambiguous trailing zero (no decimal). Try 2500 vs 2500. vs 2.5e3.

03

Solved Examples (Step-by-Step)

Count → rule → round. Jo teen moves yahan hain, wahi har sig-fig MCQ ka poora game hai.

EXAMPLE 01Foundation · Counting

Four numbers, four rulings

State the number of significant figures in: 2.50, 0.0045, 2005, 0.02050.

  1. 2.50Decimal present → trailing zero counts → 3 s.f.
  2. 0.0045Leading zeros skip → 4 and 5 count → 2 s.f.
  3. 2005Middle zeros sit between non-zeros → 4 s.f.
  4. 0.02050Leading skip; then 2, 0, 5 and final trailing 0 (after decimal) → 4 s.f.

3 · 2 · 4 · 4 significant figures

EXAMPLE 02JEE Main · × ÷ rule

Multiplication under the least-sig-fig law

Express the result of 4.20 × 1.4 × 3.05 with proper significant figures.

  1. Raw product4.20 × 1.4 × 3.05 = 17.934
  2. Governing inputs.f. counts: 4.20 → 3, 1.4 → 2, 3.05 → 3 — fewest = 2.
  3. Round17.934 → 18 — write 1.8 × 10¹ to make the 2 s.f. visible.
  4. Check“18” alone is ambiguous (trailing zeros rule) — notation rescues the precision claim ✓

1.8 × 10¹ (2 significant figures)

EXAMPLE 03JEE Main · + − rule & half-even

Addition under the least-decimal law, plus a tie

Evaluate 3.9 + 4.52 + 2.005 with correct precision, and round 2.45 to 2 significant figures.

  1. Raw sum3.9 + 4.52 + 2.005 = 10.425
  2. Governing inputDecimal places: 3.9 → 1, 4.52 → 2, 2.005 → 3 — fewest = 1.
  3. Round10.425 → 10.4 (1 decimal place) — places, not sig figs.
  4. The tie2.45 → 2 s.f.: digit after is exactly 5 → round to even → keep 4 → 2.4. (2.4501 → 2.5, since something follows the 5.)

10.4 · 2.4 (half-even tie resolved downward)

05

Key Takeaways

Rule card

Eight lines that solve this topic

Non-zero digits: always s.f.425 → 3 s.f., no debate.
Zeros between non-zeros: count2005 → 4 s.f.; 0.02050’s middle zero counts too.
Leading zeros: never count0.0034 → 2 s.f. — they’re placeholder grammar.
Trailing zeros: count iff decimal present2.500 → 4; 2500 → ambiguous → 2.5 × 10³.
Exact numbers: unlimited s.f.Counts and defined factors (100 cm = 1 m) never limit.
× ÷ → fewest s.f.4.2 × 1.40 = 5.88 → 5.9 (2 s.f.).
+ − → fewest decimal places3.9 + 4.52 = 8.42 → 8.4 (1 dp). Places, not digits.
Exactly 5, alone → round to even2.45 → 2.4 (2 s.f.); 2.4501 → 2.5.

Notation anchors: 0.02050 = 2.050 × 10⁻² (4 s.f.) · 3200 g → 3.2 × 10³ g (2 s.f.) · 1 u = 1.6605 × 10⁻²⁴ g (5 s.f.)  ·  precision = spread · accuracy = truth

  1. Sig figs are a precision contract — the last digit you write is the uncertainty you admit.
  2. The decimal point decides trailing zeros — present they count, absent they’re ambiguous; scientific notation ends the argument.
  3. Two operation rules, never swapped — × ÷ counts significant figures, + − counts decimal places.
  4. Precision ≠ accuracy — tight clusters can be systematically wrong; FIG. 1’s target grid is the one-glance cure.
06

FAQs

What are significant figures and why do they matter?

Significant figures are all the digits in a measured value that are known reliably plus the first uncertain digit — they report how precise a measurement is. Writing 2.50 g rather than 2.500 g claims different precision. More significant figures mean a more precise measurement, and calculated answers may not exceed the precision of the least precise input.

What are the rules for counting significant figures?

Non-zero digits are always significant; zeros between non-zero digits are significant (2005 → 4); leading zeros are never significant (0.0034 → 2); trailing zeros are significant only when a decimal point is present (2.500 → 4, but 2500 is ambiguous — write 2.5 × 10³). Exact numbers, such as counted objects or defined factors like 100 cm = 1 m, have unlimited significant figures.

What is the difference between accuracy and precision?

Accuracy is how close a measurement is to the true value; precision is how close repeated measurements are to each other. A balance reading 5.000, 5.001 and 4.999 g for a true 5 g mass is both precise and accurate; readings of 4.890, 4.891 and 4.889 g are precise but inaccurate.

How many significant figures should the answer to a calculation have?

In multiplication and division, the answer carries the smallest number of significant figures among the inputs — 4.2 × 1.40 = 5.88 → 5.9 (2 significant figures). In addition and subtraction, the answer is rounded to the smallest number of decimal places — 3.9 + 4.52 = 8.42 → 8.4 (1 decimal place).

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QCC Notes — Class 11 Chemistry

Strictly NCERT-aligned notes for JEE Main & NEET, prepared by QCC Notes (Padho Likho JEE). Content follows the latest NCERT edition and current NTA exam pattern.

Last updated
30 Aug 2026