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NEET Chemistry MCQ
Quiz 1
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Q.1
The value of $\frac {1.0042 \times .0034}{1.23}$ using significant figure is
0%
.0028
0%
.003
0%
.002
0%
.00279
Explanation
1.0042 (5 sig figs) × 0.0034 (2 sig figs) → intermediate result limited to 2 sig figs; dividing by 1.23 (3 sig figs) keeps the final answer to 2 sig figs (the smallest count involved). (1.0042 × 0.0034) / 1.23 = 0.0034143/1.23 ≈ 0.0027758 → rounded to 2 significant figures: 0.0028.
Q.2
841.45 to four significant figures will be
0%
841.0
0%
841.40
0%
841.5
0%
841.4
Explanation
841.45 rounded to 4 significant figures: the digit right after the 4th significant figure is exactly 5, and using the standard "round half to even" convention, the preceding digit (4, already even) stays as is: 841.4.
Q.3
Which is of these zero is not significant.
0%
.901
0%
.011
0%
.1240
0%
.309
Explanation
In 0.011, the leading zero (before the first '1') is not significant — it's just a placeholder marking the decimal position. (In 0.901, 0.1240, and 0.309, the zeros sit between or after nonzero digits and ARE significant.)
Q.4
The number of Moles of KCL is 1000 ml of 3 Molar Solution is
0%
1
0%
1.5
0%
3
0%
2
Explanation
Moles = Molarity × Volume(L) = 3 mol/L × 1.000 L = 3 mol.
Q.5
we mix three solution of same solute of Molarity 1M, 2M ,3M and Volume 1 L,2L ,3L. Find the molarity of the resultant solution Molarity of the resulting solution is given by
0%
3 M
0%
3.3 M
0%
2.33 M
0%
1 M
Explanation
Total moles = (1M×1L) + (2M×2L) + (3M×3L) = 1 + 4 + 9 = 14 mol. Total volume = 1 + 2 + 3 = 6 L. Molarity = 14/6 ≈ 2.33 M.
Q.6
Which of the following modes of expressing concentration is independent of temperature
0%
Molality
0%
Formality
0%
Normality
0%
Molarity
Explanation
Molality is based on mass of solvent (kg), which doesn't change with temperature — unlike molarity, normality, and formality, which are all based on the solution's VOLUME, and volume expands or contracts with temperature.
Q.7
If 100 ml of 1.0 M NaOH solution is diluted to 1 L, the resulting solution contains
0%
.1 Mole of NaOH
0%
.05 Mole of NaOH
0%
1 Mole of NaOH
0%
10 mole of NaOH
Explanation
Dilution changes concentration but never changes the number of moles present. Moles of NaOH = Molarity × Volume(L) = 1.0 × 0.100 = 0.1 mol — this stays the same after dilution to 1 L, just at a lower concentration.
Q.8
A Molal solution is one that contains one mole of solute in
0%
1 L of solvent
0%
1000 g of solvent
0%
22.4 L of solution
0%
1 L of solution
Explanation
By definition, a molal (1 m) solution contains exactly one mole of solute dissolved in 1000 g (1 kg) of solvent.
Q.9
Assertion (A) : Significant figures for 0.10 is 2 where as for 10 it isReason (R) : Zero at the end or right of a number are significant provided they are not on the right side of the decimal point.
0%
A is true but R is false.
0%
Both A and R are true and R is correct explanation of A.
0%
Both A and R are false.
0%
Both A and R are true but R is not a correct explanation of A.
Explanation
0.10 has 2 significant figures (the trailing zero after the decimal point counts), while plain 10 (no decimal point shown) is conventionally read as having just 1 significant figure — so the Assertion is correct. But the Reason has the rule backwards: trailing zeros ARE significant specifically when they DO appear after a decimal point, not when they don't — the Reason states the opposite of the correct rule, making it false.
Q.10
Assertion (A) : One atomic mass unit is defined as one twelfth of the mass of one carbon-12 atom. Reason (R) : Carbon-12 isotope is the most abundunt isotope of carbon and has been chosen as standard.
0%
Both A and R are false.
0%
Both A and R are true but R is not the correct explanation of A.
0%
A is true but R is false.
0%
Both A and R are true and R is the correct explanation of A.
Explanation
Both statements are true on their own — one atomic mass unit genuinely is defined as 1/12 the mass of a carbon-12 atom, and carbon-12 genuinely is the most abundant carbon isotope (about 98.9% of natural carbon). But the choice of carbon-12 as the atomic-mass reference standard wasn't made specifically BECAUSE of its abundance — it was chosen mainly for the practical convenience of giving other elements' atomic masses close to whole numbers. So the Reason doesn't correctly explain the Assertion.
Q.11
The mole fraction of solvent in a solution where mole fraction of solute is .6
0%
.4
0%
1.6
0%
.6
0%
1
Explanation
Mole fractions of all components in a solution always add up to 1. So mole fraction of solvent = 1 − 0.6 = 0.4.
Q.12
The value of $2.01 \times .011$ using significant figure is
0%
.023
0%
.022
0%
.02211
0%
.0222
Explanation
2.01 has 3 significant figures, 0.011 has 2 (leading zeros don't count) — for multiplication, the result keeps the smaller count, 2 sig figs. 2.01 × 0.011 = 0.02211 → rounded to 2 significant figures: 0.022.
Q.13
The value of $5396 \times .045 + 325.3$ is
0%
565
0%
568.3
0%
568
0%
560
Explanation
5396 (4 sig figs) × 0.045 (2 sig figs) → result limited to 2 sig figs: 5396 × 0.045 = 242.82 → 240 (2 sig figs). Adding 240 + 325.3: for addition, the result is limited to the least number of decimal places among the terms — 240 has 0 decimal places, so the sum is rounded to 0 decimal places: 240 + 325.3 = 565.3 → 565.
Q.14
The value of 70.3 - 1.245 using significant figure is
0%
69.1
0%
69.06
0%
69.055
0%
69.05
Explanation
For subtraction, the result keeps only as many decimal places as the term with the FEWEST decimal places. 70.3 has 1 decimal place, 1.245 has 3 — so the result is rounded to 1 decimal place. 70.3 − 1.245 = 69.055 → 69.1.
Q.15
The value of 10.5+1.51+2.401 using significant figure is
0%
14.5
0%
14.41
0%
14.411
0%
14.4
Explanation
For addition, the result keeps only as many decimal places as the term with the fewest — 10.5 has 1 decimal place (the fewest of the three), so the sum is rounded to 1 decimal place. 10.5 + 1.51 + 2.401 = 14.411 → 14.4.
Q.16
Match the column
0%
p -> i, q -> iii, r -> iv, s-> ii
0%
p -> i, q -> iii, r -> ii, s-> iv
0%
p -> ii, q -> iii, r -> ii, s-> i
0%
p -> ii, q -> iii, r -> i, s-> ii
Explanation
Counting significant figures for each: 0.0015 → 2 (leading zeros don't count); 0.0105 → 3 (the zero between 1 and 5 is a captive zero and does count); 1.500 → 4 (trailing zeros after a decimal point count); 15.000 → 5 (same rule, all five digits count). (Note: Column II's own listing repeats the label "(ii)" for two different rows in the source image, which explains why the marked answer's lettering looks inconsistent — the underlying values above are the reliable, correct part.)
Q.17
The scientific notation for the number 0.00000060 is
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$6 \times 10^{-7}$
0%
$6.0 \times 10^{-7}$
0%
$6.00 \times 10^{-7}$
0%
$60 \times 10^{-8}$
Explanation
0.00000060 is written with a trailing zero after the 6, meaning it's specified to 2 significant figures — so its scientific notation must also show exactly 2 significant figures: 6.0 × 10⁻⁷ (not 6 × 10⁻⁷, which would only show 1, or 6.00 × 10⁻⁷, which would show 3).
Q.18
The significant figures in the number 1200 is
0%
2
0%
5
0%
3
0%
4
Explanation
Without a decimal point or other explicit marker, trailing zeros in a whole number like 1200 are treated as not significant by default — so it has only 2 significant figures (the '1' and the '2').
Q.19
70.770 to three significant figures will be
0%
70.7
0%
70.0
0%
71.0
0%
70.8
Explanation
70.770 rounded to 3 significant figures: keeping the first three digits (7, 0, 7) and looking at the next digit (7, which rounds up): 70.8.
Q.20
which of these is not having 3 significant figures
0%
11100
0%
4.01
0%
1.410
0%
1.00
Explanation
1.410 has a decimal point with a trailing zero after it, so all four digits (1, 4, 1, 0) count — that's 4 significant figures, not 3. (11100, without a decimal point, is conventionally read as 3 sig figs; 4.01 has 3 sig figs since its zero is captive; and 1.00 has 3 sig figs since the trailing zeros after the decimal count.)
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