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NEET Chemistry MCQ
Quiz 2
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Q.1
12.1756 to 5 significant figures
0%
12.176
0%
12.180
0%
12.175
0%
12.170
Explanation
12.1756 rounded to 5 significant figures: keeping the first five digits (1,2,1,7,5) and looking at the 6th digit (6, which rounds the 5 up to 6): 12.176.
Q.2
The value of $1.23 \times .231$ is
0%
.280
0%
0.285
0%
0.284
0%
0.28413
Explanation
1.23 and 0.231 both have 3 significant figures, so the product is rounded to 3 sig figs too. 1.23 × 0.231 = 0.28413 → 0.284.
Q.3
The value of 415.5 + 3.64 + .238 is
0%
419.378
0%
419.4
0%
419.38
0%
420
Explanation
For addition, the result is limited to the fewest decimal places among the terms — 415.5 has 1 decimal place (the fewest), so the sum rounds to 1 decimal place. 415.5 + 3.64 + 0.238 = 419.378 → 419.4.
Q.4
A solution of glucose in water is labelled as 10% w/w, The molality is
0%
.74 moles/kg
0%
.5 moles/kg
0%
0.62 moles/kg
0%
none of these
Explanation
A 10% w/w solution means 10 g glucose per 100 g of solution, i.e. per 90 g of water. Moles of glucose (molar mass 180 g/mol) = 10/180 ≈ 0.0556 mol Molality = moles solute / kg solvent = 0.0556 / 0.090 ≈ 0.62 mol/kg.
Q.5
Which of the following terms are unit less?
0%
Mole fraction
0%
Formality
0%
Molality
0%
Molarity
Explanation
Mole fraction — the ratio of moles of one component to total moles — is a pure ratio and carries no units. Formality, molality and molarity all have units (like mol/kg or mol/L).
Q.6
Molarity of Solution having 5 moles of solutes present in 2 L of solution is
0%
2.5 M
0%
4 M
0%
.5 M
0%
2 M
Explanation
Molarity = moles of solute / volume of solution (in litres) = 5 mol / 2 L = 2.5 M.
Q.7
The correct expression relating molality(m) , Molarity(M), density (d) and Molar Mass(MB) of solute is
0%
$m = \frac { d - MM_B }{ M}$
0%
$m = \frac {M}{d - MM_B}$
0%
$m = \frac {M}{d + MM_B}$
0%
$m = \frac { d + MM_B }{M }$
Explanation
Deriving the relation for 1 L of solution: mass of solution = 1000d grams (d in g/mL), moles of solute = M (from the definition of molarity), so mass of solute = M×M_B grams, and mass of solvent = (1000d − M·M_B) grams = (1000d − M·M_B)/1000 kg. Molality = moles solute / kg solvent = 1000M / (1000d − M·M_B). (The marked option here appears to be missing the "1000" factors — likely lost during scraping — but this is the standard, correct relation connecting m, M, d, and M_B, and matches the marked form once consistent units absorb that constant.)
Q.8
The molality of the solution which contains 10 g of Cane sugar($C_{12}H_{22}O_{11}$) dissolved in 150 g of water?
0%
.22 moles/kg
0%
1.1 moles/kg
0%
.5 moles/kg
0%
.194 moles/kg
Explanation
Cane sugar (C₁₂H₂₂O₁₁) has molar mass 342 g/mol. Moles = 10/342 ≈ 0.0292 mol Molality = moles/kg solvent = 0.0292 / 0.150 ≈ 0.194 mol/kg.
Q.9
How many moles of NaCl are contained in 100.0 mL of a 0.200 M solution?
0%
.0002
0%
.002
0%
2
0%
.02
Explanation
Moles = Molarity × Volume(L) = 0.200 mol/L × 0.100 L = 0.02 mol.
Q.10
find the molarity of the solution where mole fraction of solute is 0.5000 and Molar Mass of Solute is 49 g/Mole and Molar mass of solvent is 18 g/M and density of solution is 1.769 g/ml
0%
15.24
0%
12.24
0%
11.13
0%
10.1
Explanation
Using the general method: with mole fraction of solute = 0.5, take 1 mol solute and 1 mol solvent (equal moles, matching a 0.5 mole fraction). Mass of solute = 1 × 49 = 49 g; mass of solvent = 1 × 18 = 18 g; total mass = 67 g. Volume = mass/density = 67/1.769 ≈ 37.87 mL = 0.03787 L Molarity = moles solute/volume = 1/0.03787 ≈ 26.4 M. (This standard-method calculation doesn't match the marked 15.24 M using the numbers exactly as given — this looks like it may be a fragment of a longer, multi-part problem that lost some connecting context during scraping. The method above is the correct general approach for this type of mole-fraction-to-molarity conversion regardless.)
Q.11
What will be the molarity of a solution, which contains 5.85 g of NaCl per 500ml?
0%
.5 M
0%
.4 M
0%
1 M
0%
.2 M
Explanation
NaCl has molar mass 58.5 g/mol. Moles = 5.85/58.5 = 0.1 mol Molarity = moles/volume(L) = 0.1/0.500 = 0.2 M.
Q.12
What is the molarity of a solution that contains 1.724 moles of solute in 2.50 L of solution?
0%
.55 M
0%
1.1 M
0%
.670 M
0%
.690 M
Explanation
Molarity = moles of solute / volume of solution = 1.724 mol / 2.50 L ≈ 0.690 M.
Q.13
The density of a 2.03 M solution of Acetic acid is water is 1.017 g/mL. The molality of the solution is
0%
2.27 m
0%
1.52 m
0%
2.5 m
0%
2 m
Explanation
Acetic acid (CH₃COOH) has molar mass 60 g/mol. For 1 L of solution: Mass of solution = 1.017 g/mL × 1000 mL = 1017 g Moles of solute = 2.03 mol (from the molarity), so mass of solute = 2.03 × 60 = 121.8 g Mass of solvent = 1017 − 121.8 = 895.2 g = 0.8952 kg Molality = 2.03 / 0.8952 ≈ 2.27 mol/kg.
Q.14
Which of the following pairs have the same number of atoms?
0%
28 g of N2 and 32 g of O2
0%
16 g of O2 and 44 g of CO2
0%
16 g of O2(g) and 4 g of H2(g)
0%
12 g of C(s) and 46 g of Na(s)
Explanation
Moles of N₂ in 28 g = 28/28 = 1 mol → 2 mol atoms (N₂ has 2 atoms per molecule). Moles of O₂ in 32 g = 32/32 = 1 mol → 2 mol atoms (O₂ also has 2 atoms per molecule). Both give exactly 2 mol of atoms — a match. (Checking the others: 16 g O₂ gives 1 mol atoms vs 44 g CO₂ giving 3 mol atoms; 16 g O₂ gives 1 mol atoms vs 4 g H₂ giving 4 mol atoms; 12 g C gives 1 mol atoms vs 46 g Na giving 2 mol atoms — none of these match.)
Q.15
The value of $914 + 23.2$ using significant figures is
0%
930
0%
937.2
0%
938
0%
937
Explanation
For addition, the result keeps only as many decimal places as the term with the fewest — 914 has 0 decimal places, so the sum is rounded to a whole number. 914 + 23.2 = 937.2 → 937.
Q.16
The number of gram molecule of oxygen in $6.02 \times 10^{24}$ CO molecules is
0%
5 gm-molecules
0%
.5 gm-molecules
0%
10 gm-molecules
0%
1 gm-molecules
Explanation
6.02×10²⁴ CO molecules = 6.02×10²⁴ / 6.022×10²³ = 10 mol of CO molecules. Each CO molecule contains 1 oxygen atom, so that's 10 mol of oxygen ATOMS — which corresponds to 10/2 = 5 gram-molecules (moles) of O₂, since a gram-molecule of oxygen means moles of O₂ molecules, each containing 2 oxygen atoms.
Q.17
The number of significant figures in 1060.50 is
0%
4
0%
3
0%
6
0%
5
Explanation
1060.50 is written with an explicit decimal point and a trailing zero after it, so every digit shown counts as significant: 1, 0, 6, 0, 5, 0 — 6 significant figures.
Q.18
The number of atoms of oxygen present in 88g of $CO_2$ is
0%
$24.092 \times 10^{22}$
0%
$12.046 \times 10^{23}$
0%
$24.092 \times 10^{23}$
0%
$6.023 \times 10^{23}$
Explanation
CO₂ has molar mass 44 g/mol. Moles of CO₂ in 88 g = 88/44 = 2 mol. Each CO₂ molecule has 2 oxygen atoms, so moles of O atoms = 2 × 2 = 4 mol. Number of O atoms = 4 × 6.023×10²³ = 24.092×10²³.
Q.19
In the reaction 2X + 4Y → 3U + 4V, when 5 moles of X react with 6 moles of Y, Then which of the following is incorrect
0%
the amount of V formed will be 5 moles
0%
the amount of U formed is 4.5 moles
0%
Insufficient information
0%
‘Y’ is the limiting reagent.
Explanation
For 2X + 4Y → 3U + 4V, the required ratio is 1 mol X to 2 mol Y. With 5 mol X, all of it would need 10 mol Y — but only 6 mol Y is available, so Y is the limiting reagent (confirming that statement is true), and only 3 mol of X (6 mol Y × 1/2) actually reacts. Using the limiting reagent (6 mol Y): U formed = 6 × (3/4) = 4.5 mol (matches the given option), and V formed = 6 × (4/4) = 6 mol — NOT 5 mol as claimed, which makes "V formed will be 5 moles" the incorrect statement.
Q.20
Match the column
0%
p -> i, q -> iii, r -> ii, s-> iv
0%
p -> ii, q -> iv, r -> iii, s-> i
0%
p -> ii, q -> iii, r -> iv, s-> i
0%
p -> ii, q -> i, r -> iv, s-> iii
Explanation
Matching each physical quantity to its SI unit: (p) Luminous intensity → Candela (ii) (q) Pressure → Pascal (iii) (r) Molality → mole/kg (iv) (s) Density → kg/m³ (i) So: p → ii, q → iii, r → iv, s → i.
0 h : 0 m : 1 s
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