CBSE Class 11 Chemistry – MCQ and Online Tests – Unit 1 – Some Basic Concepts of Chemistry
Every year CBSE schools conducts Annual Assessment exams for 6,7,8,9,11th standards. These exams are very competitive to all the students. So our website provides online tests for all the 6,7,8,9,11th standard’s subjects. These tests are also very effective and useful for those who preparing for any competitive exams like Olympiad etc. It can boost their preparation level and confidence level by attempting these chapter wise online tests.
These online tests are based on latest CBSE syllabus. While attempting these our students can identify the weak lessons and continuously practice those lessons for attaining high marks. It also helps to revise the NCERT textbooks thoroughly.
CBSE Class 11 Chemistry – MCQ and Online Tests – Unit 1 – Some Basic Concepts of Chemistry
Question 1.
The significant figures in 3400 are
(a) 2
(b) 5
(c) 6
(d) 4
Answer
Answer: (b) 5
Explanation:
As we know that all non-zero unit are significant number.
Question 2.
The total number of ions present in 111 g of CaCl2 is
(a) One Mole
(b) Two Mole
(c) Three Mole
(d) Four Mole
Answer
Answer: (c) Three Mole
Explanation:
Molecular weight of
CaCl2 = 111g/mol
Ions in one calcium chloride molecule = Ca+2 + 2Cl– = 3 ions
Now no. of molecules in 111g/mol of CaCl2 = Avogadros number
= 6.02 × 1023 molecules
So number of ions in 111g/mol of CaCl2
= 3 × 6.02 × 1023 ions
= 3 moles.
Question 3.
What is the concentration of nitrate ions if equal volumes of 0.1 M AgNO3 and 0.1 M NaCl are mixed together
(a) 0.1 N
(b) 0.2 M
(c) 0.05 M
(d) 0.25 M
Answer
Answer: (b) 0.2 M
Explanation:
0.1 M AgNOA will react with 0.1 M NaCl to form 0.1 M NaNOA.
But as the volume is doubled, conc. of NO–3 = \(\frac { (0.1) }{ (2) } \)
= 0.05 M
Question 4.
The -ve charged particles is called:
(a) Anion
(b) Cation
(c) Radical
(d) Atom
Answer
Answer: (a) Anion
Explanation:
A charged particle, also called an ion, is an atom with a positive or negative charge.
This happens whenever something called an ionic bond forms.
Two particles that have different numbers of electrons (the smallest particle in an atom which is negative) start reacting to each other.
The particle that has the greater amount of electrons takes the other particles electrons.
One becomes positive because it lost an electron, and the other negative because it got another electron.
The two particles become attracted to each other and mix together, making a new kind of particle.
Question 5.
Which is not a unit of pressure:
(a) Bar
(b) N/m²
(c) Kg/m²
(d) Torr
Answer
Answer: (c) Kg/m²
Explanation:
Pressure is force per unit area so Kg/m² cannot be a unit of pressure.
Question 6.
What is the normality of a 1 M solution of H3PO4
(a) 0.5 N
(b) 1.0 N
(c) 2.0 N
(d) 3.0 N
Answer
Answer: (d) 3.0 N
Explanation:
H3PO4 is tribasic
So N = 3M
= 3 × 1 = 3.
Question 7.
Which of the following weighs the most?
(a) One g – atom of nitrogen
(b) One mole of water
(c) One mole of sodium
(d) One molecule of H2SO4
Answer
Answer: (c) One mole of sodium
Question 8.
Under similar conditions of pressure and temperature, 40 ml of slightly moist hydrogen chloride gas is mixed with 20 ml of ammonia gas, the final volume of gas at the same temperature and pressure will be
(a) 100 ml
(b) 20 ml
(c) 40 ml
(d) 60 ml
Answer
Answer: (b) 20 ml
Explanation:
NH3(g) + HCl(g)? NH4Cl(s)
t = 0 40ml 0
t = t 20ml solid
Final volume = 20ml.
Question 9.
An organic compound contains carbon , hydrogen and oxygen. Its elemental analysis gave C, 38.71% and H, 9.67%. The empirical formula of the compound would be
(a) CHO
(b) CH4O
(c) CH3O
(d) CH2O
Answer
Answer: (c) CH3O
Explanation:
Question 10.
Which of the following cannot give iodometric titrations
(a) Fe3+
(b) Cu2+
(c) Pb2+
(d) Ag+
Answer
Answer: (c) Pb2+
Explanation:
Atom in highest oxidation state can oxidize iodide to liberate I2 which is volumetrically measured by iodometric titration using hypo.
2I– ?I2
Pb+2 ? Lowest oxidation state cannot oxidise iodide to I2
Question 11.
Which of the following contains same number of carbon atoms as are in 6.0 g of carbon (C – 12)?
(a) 6.0 g Ethane
(b) 8.0 g Methane
(c) 21.0 g Propane
(d) 28.0 g CO
Answer
Answer: (b) 8.0 g Methane
Explanation:
6g carbon
Moles of carbon = (6/12) = 0.5 mol
Number of carbon atoms
= 0.5 × NA =0.5NA (NA is Avogadro number)
6g ethane (C2H6 two atoms of C per mole)
Moles = (6/30) = 0.2 mol
Number of carbon atoms = 0.2 × 2 × NA = 0.4 NA
(Number of carbon atoms = moles of compound × number of C atoms per mol × Avogadro number)
8g methane (CH4)
Moles = (8/16) = 0.5 mol
Number of carbon atoms = 0.5 × 1 × NA = 0.5 NA
21g propane (C3H8)
Moles = (21/44) =0.48 mol
Number of carbon atoms = 0.48 × 3 × NA = 1.44 NA
28g CO
Moles = (28/28) =1 mol
Number of carbon atoms = 1 × 1 × NA = NA
Question 12.
Approximate atomic weight of an element is 26.89. If its equivalent weight is 8.9, the exact atomic weight of element would be
(a) 26.89
(b) 8.9
(c) 17.8
(d) 26.7
Answer
Answer: (d) 26.7
Explanation:
Atomic weight = (Equivalent weight × Valency)
=(8.9 × 3) = 26.7
(Valency = (26.89)/(8.9) ˜ 3).
Question 13.
The number of moles present in 6 gms of carbon is:
(a) 2
(b) 0.5
(c) 5
(d) 1
Answer
Answer: (b) 0.5
Explanation:
The molar mass of
12C is 12.0 gmol-1.
NA (Avogadros number = 6.022×1023)
12C atoms have a mass of 12.0 g.
Given that: – 6.0 g.
Thus (6.0 g)/ (12.0 gmol-1) = 0.50 mol.
Question 14.
The sulphate of a metal M contains 9.87% of M. This sulphate is isomorphous with ZnSO4.7H2O. The atomic weight of M is
(a) 40.3
(b) 36.3
(c) 24.3
(d) 11.3
Answer
Answer: (c) 24.3
Explanation:
As the given sulphate is isomorphous with ZnSO4.7H2O its formula would be MSO4.7H2O.m is the atomic weight of M, molecular weight of MSO4.7H2O
= (m + 32 + 64 + 126) = (m + 222)
Hence % of M= [(m)/ (m+222)] × 100 = 9.87(given)
Or
100 m = (9.87 m + 222 × 9.87)
Or 90.13 m = (222 × 9.87)
Or m = (222 × 9.87)/(90.13)
= 24.3.
Question 15.
The S.I unit of temperature is :
(a) Kelvin
(b) Celsius
(c) Fahrenheit
(d) Centigrade
Answer
Answer: (a) Kelvin
Explanation:
The SI unit of temperature is the Kelvin. Although the Celsius temperature scale is also used, it is considered a derived SI unit and is generally used to measure everyday temperatures.
In the Kelvin scale, the lowest possible temperature is 0 K. This reading is called absolute zero; however, nothing could possibly ever reach this temperature. Absolute zero, according to the Celsius scale, is approximately minus 273 degrees Celsius. Water boils at 373 Kelvin and freezes at 273 Kelvin.
Scientists discovered absolute zero when they figured that as temperature decreases, the volume of a gas also gets smaller. They graphed this relationship and found that for each substance tested, zero volume for each substance would hypothetically occur around minus 273 degrees Celsius, the equivalent of absolute zero.
Question 16.
A symbol not only represents the name of the element but also represents
(a) Atomic Mass
(b) Atomic Number
(c) Atomicity
(d) Atomic Volume
Answer
Answer: (c) Atomicity
Explanation:
Symbol of an element represents the name along with the number of same atoms in room condition, which is its atomicity, of that element.
Question 17.
What is the normality of a 1 M solution of H3PO4
(a) 0.5 N
(b) 1.0 N
(c) 2.0 N
(d) 3.0 N
Answer
Answer: (d) 3.0 N
Explanation:
H3PO4 is tribasic
So N = 3M
= 3 × 1 = 3.
Question 18.
Which of the following weighs the most?
(a) One g – atom of nitrogen
(b) One mole of water
(c) One mole of sodium
(d) One molecule of H2SO4
Answer
Answer: (c) One mole of sodium
Question 19.
What is the concentration of nitrate ions if equal volumes of 0.1 MAgNO3 and 0.1 M NaCl are mixed together
(a) 0.1 M
(b) 0.2 M
(c) 0.05 M
(d) 0.25 M
Answer
Answer: (c) 0.05 M
Explanation:
0.1M AgNO3 will react with 0.1M NaCl to form 0.1M NaNO3.
But as the volume doubled, conc. of NO–2
= 0.12
= 0.05 M
Question 20.
The number of moles present in 6 gms of carbon is:
(a) 2
(b) 0.5
(c) 5
(d) 1
Answer
Answer: (b) 0.5
Explanation:
The molar mass of
12C is 12.0 gmol-1.
NA (Avogadros number = 6.022 × 1023)
12C atoms have a mass of 12.0 g.
Given that: – 6.0 g.
Thus (6.0 g)/ (12.0 gmol-1) = 0.50 mol.
0 comments:
Post a Comment