11th Standard Syllabus & Materials
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Published on: 01/08/2019
Download Tamil Nadu 11th Standard Chemistry question papers, model tests, one-mark questions, important questions, and public exam papers in PDF format. Free study materials and answer keys for TN State Board students.
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1.
Molecular mass =
Vapour Density x 2
Vapour Density \(\div\)2
Vapour Density x 3
Vapour Density
2.
3.
Hot concentrated sulphuric acid is a moderately strong oxidizing agent. Which of the following reactions does not show oxidising behaviour ?
Cu + 2H2 SO4 \(\longrightarrow \) CuSO4 +SO2 + 2H2O
C + 2H2 + SO4 \(\longrightarrow \) CO2 + 2SO2 + 2H2O
BaCl2 + H2SO4 \(\longrightarrow \) BaSO4 + 2HCl
None of the above
4.
40 ml of methane is completely burnt using 80 ml of oxygen at room temperature The volume of gas left after cooling to room temperature is _______.
40 ml CO2 gas
40 ml CO2 gas and 80 ml H2O gas
60 ml CO2 gas and 60 ml H2O gas
120 ml CO2 gas
5.
State Mendeleev's periodic law.
6.
7.
Give the number of electrons in the following species. H2 , H2 , O2 and O2-
8.
What do you understand by the term oxidation number ?
9.
In the below figure, let us find the missing parameters [volume in (b) and temperature in (c)]
PI = 1 atm, P2 = 1 atm, P3 = 1 atm
V1 = 0.3 dm3, V2 = ? dm3, V3 = 0.15 dm3
T1 = 200 K, T2 = 300 K, T3 = ? K.

10.
What are inner transition elements?
11.
Calculate the Formula Weights of the following compounds. NaOH
12.
What is the difference between molecular mass and molar mass ? Calculate the molecular mass and molar mass for carbon monoxide.
13.
How fast must a 54g tennis ball travel in order to have a de Broglie wavelength that is equal to that of a photon of green light 5400\(\overset { 0 }{ A } \) ?
14.
An atom of an element has 19 electrons. What is the total number of p-orbital?
15.
Explain the merits of Moseley's long form of periodic table.
16.
A helium filled balloon had a volume of 400 mL, when it is cooled to -120°C. what will be its volume if the balloon is warmed in an oven to 100°C assuming changes in pressure.
17.
Distinguish between the following.
(i) Atomic and molecular mass
(ii) Atomic mass and atomic weight
(iii) Empirical and molecular formula
(iv) Moles and molecules.
18.
The reaction between aluminium and ferric oxide can generate temperatures up to 3273 K and is used in welding metals. (Atomic mass of Al = 27 u atomic mass of O = 16 u )
2Al + Fe2O3 \(\longrightarrow \) Al2O3 + 2Fe; If in this process, 324 g of aluminum is allowed to react with 1.12 kg of ferric oxide
i) Calculate the mass of Al2O3 formed
ii) How much of the excess reagent is left at the end of the reaction ?
19.
Assertion: Helium has the highest value of ionization energy among all the elements known
Reason: Helium has the highest value of electron affinity among all the elements known.
Codes:
(a) Both assertion and reason are true and reason is correct explanation for the assertion.
(b) Both assertion and reason are true but the reason is not the correct explanation for the assertion
(c) Assertion is true and the reason is false
(d) Both assertion and the reason are false
Both assertion and reason are true and reason is correct explanation for the assertion.
Both assertion and reason are true but the reason is not the correct explanation for the assertion
Assertion is true and the reason is false
Both assertion and the reason are false
1.
(a)
Vapour Density x 2
2.
(a)
3.
(c)
BaCl2 + H2SO4 \(\longrightarrow \) BaSO4 + 2HCl
4.
(a)
40 ml CO2 gas
5.
This law states that "The physical and chemical properties of elements are a periodic function of their atomic weights."
6.
7.
H2 = 1 + 1 = 2e-
H+2; = 2 - 1 = le-
\({O}_{2}\) = 8 + 8 = 16e-
\({O}_{2}^{-}\) = 8 + 8 + 1= 17e-
8.
It is defined as the imaginary charge left on the atom when all other atoms of the compound have been removed in their usual oxidation states that are assigned according to set of rules.
9.
According to Charles' law,
\(\frac{V_1}{T_1}=\frac{V_2}{T_2}=\frac{V_3}{T_3}\)
\(\frac{0.3dm^3}{200K}=\frac{V_2}{300K}=\frac{0.15dm^3}{T_3}\)
\(\frac{V_2}{300K}=\frac{0.3dm^3}{200K}\)
.png)
V2 = 0.45 dm3 and
\(\frac{0.15dm^3}{T_3}=\frac{0.3dm_3}{200K}\)
.png)
T3 = 100 K.
10.
The elements in which extra electron enters (n-2) f-orbitals are called f-block elements. These elements form a transition series within the transition elements. So these elements are called inner transition elements.
11.
1 x AW of Na = 1 x 22.99 = 22.99 amu
1 x AW of O = 1x 16 = 16.00 amu
1 x AW of H = 1 x1.008 = 1.008 amu
Formula weight of NaOH is = 39.998 amu
12.
1) The unit of molecular mass is atomic mass unit [amu]. The unit of molar mass is gram per mole.
2) Molecular mass is the mass of one molecule while molar mass is the mass of one mole of molecules (6.022 x 1023)
(i) Molecular mass of CO2 = 1(C) + 2(0) = 12 + 32 = 44 amu
or 7.304 x 10-23 g
(ii) Molar mass of CO2 = 44 g mol-1.
13.
De Broglie wavelength of the tennis ball equal to 5400 \(\overset { 0 }{ A } \).
m = 54 g
V = ?
\(\lambda=\frac{h}{mV}\)
\(V=\frac{h}{m\lambda}\)
\(\mathrm{v}=\frac{6.626 \times 10^{-34} \mathrm{JS}}{54 \times 10^{-3} \mathrm{~kg} \times 5400 \times 10^{-10} \mathrm{~m}}=2.27 \times 10^{-26} \mathrm{~ms}^{-1}\)
14.
The number of electrons in an atom of an element is 19.
The electronic configuration is 1s2 2s2 2p6 3s2 3p6 4s1
The total number of p-orbitals for the element is 2p level- 3 and 3p level- 3. i.e. '6' orbitals.
15.
Merits of Moseley's long form of periodic table:
(i) As this classification is based on atomic number, it relates the position of an element to its electronic configuration.
(ii) The elements having similar electronic configuration fall in a group. They also have similar physical and chemical properties.
(iii) The completion of each period is more logical. In a period as the atomic number increases, the energy shells are gradually filled up until an inert gas configuration is reached.
(iv) The position of zero group is also justified in the table as group 18.
(v) The table completely separates metals and non-metals.
(vi} The table separates two subgroups, lanthanides and actinides, dissimilar elements do not fall together.
(vii) The greatest advantage of this periodic table is that this can be divided into four blocks namely s, p, d and f-block elements.
(viii) This arrangement of elements is easier to remember, understand and reproduce.
16.
Volume of the gas in the balloon V1 = 400 ml
temperature T1 = -120 °C + 273
= 153 K
If the balloon is warmed T2 = 100° C + 273
to a temperature = 373 K
Then the volume of the gas V2 = ?
According to charles law \(\frac { { V }_{ 1 } }{ { T }_{ 1 } } =\frac { { V }_{ 2 } }{ { T }_{ 2 } } \)
\({ V }_{ 2 }=\frac { { V }_{ 2 } }{ { T }_{ 1 } } \times { T }_{ 2 }\)
\(=\frac { 400 }{ 153 } \times 373=975\quad ml\)
\(\therefore\) Volume of helium gas in the balloon at a temperature of 100°C = 975 ml
17.
| (i) | Atomic Mass | Molecular Mass |
|---|---|---|
| Atomic mass is the mass of a single atom, which is its collective mass of neutron proton and electrons |
Molecular weight is the mass of one molecule Molecular mass can be calculated from the sum of atomic masses of all atoms present in a compound. |
|
| (ii) | Atomic Mass | Atomic Mass |
| Atomic mass is the mass of a single atom, which is its collective mass of neutron, proton and electrons |
Atomic weight is the average weight of an elements with respect to all its isotopes and their relative abundance. |
|
| (iii) | Empricial Formula | Molecular Formula |
| It represents the simplest whole number ratio of various atoms present in one molecule of the compound. Empirical formula of Benzene is CH |
The molecular formula shows the exact number of different types of atoms present in a molecule of a compound. Molecular formula of Benzene is C6H6. |
|
| (iv) | Moles | Molecules |
| The amount of the substance that contains specified particles as the number of atoms in 12 g carbon - 12 isotope |
Two or more atoms joint together by chemical bonds. |
18.
2Al + Fe2O3 \(\longrightarrow \) Al2O3 + 2Fe
| Reactants | Products | |||
| Al | Fe2O3 | Al2O3 | Fe | |
| Amount of reactant allowed to react | 324 g | 1.12 kg | - | - |
| Number of moles allowed to react | \(\frac { 324 }{ 27 } =12mol\) | \(\frac { 1.12\times { 10 }^{ 3 } }{ 160 } =7mol\) | - | - |
| Stoichiometric Co-efficient | 2 | 1 | 1 | 2 |
| Number of moles consumed during reaction | 12 mol | 6 mol | - | - |
| Number of moles of reactant unreacted and number of moles of product formed | - | 1 mol | 6 mol | 12 mol |
Molar mass of Al2O3 format = 6 mol x 102 g mol-1 = 612 g
[ Al2O3 : (2 x 27) + 3(16) = 54 + 48 = 102] = 612 g
Excess reagent = Fe2O3
Amount of excess reagent left at the end of the reaction = 1 mol x 160 g mol-1
= 160g [ Fe2O3 : (2 x 56) + (3 x 16) = 112 + 48 = 160] = 160 g
19.
(c) Assertion is true and the reason is false
11th Standard Syllabus & Materials
11th Standard
TN 11th Tamil பீடு பெற நில் - செய்யுள் - காவடிச்சிந்து Important Questions And Answers Study Material - QB365 Set A
NEW11th Standard
TN 11th Tamil பீடு பெற நில் - உரைநடை - மலை இடப்பெயர்கள் : ஓர் ஆய்வு Important Questions And Answers Study Material - QB365 Set A
NEW11th Standard
TN 11th Tamil மாமழை போற்றுதும் - துணைப்பாடம் - யானை டாக்டர் Important Questions And Answers Study Material - QB365 Set A
NEW11th Standard
TN 11th Tamil மாமழை போற்றுதும் - செய்யுள் - ஐங்குறுநூறு Important Questions And Answers Study Material - QB365 Set A
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