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Ncert Solutions Chemistry Class 11th

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11 months ago

Ncert Solutions Chemistry Class 11th The S Block Elements

10.4. Assertion: Alkali metals are always univalent.

Reason: Alkali metals are soft.
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V
Vishal Baghel

Contributor-Level 10

(b) Both the statements are correct but not the reason for the assertion.

New question posted

11 months ago

Ncert Solutions Chemistry Class 11th The S Block Elements

10.3. Assertion: The alkalimetals and their salts impart characteristiccolour to an oxidizing flame.

Reason: This heat from the flame excites the outermost orbitalelectron to a higher energy level. When the excitedelectron comes back to the ground state, thereis emission of radiation in the visible region ofthe spectrum.
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New answer posted

11 months ago

Ncert Solutions Chemistry Class 11th Some Basic Concepts of Chemistry

Assertion Type Questions:

1.37. Assertion: One mole of any gas contains 6.022 x 1023 molecules.

Reason: 1 g of any gas contains equal number of moles.
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P
Payal Gupta

Contributor-Level 10

1.37. (c) Equal moles of different substances contain same number of constituent particles but equal weights of different substances do not contain the same number of constituent particles due to the difference in their molar mass.

New answer posted

11 months ago

Ncert Solutions Chemistry Class 11th The S Block Elements

10.2. Assertion: The ionization enthalpies of the alkali metalsare considerably high and increase down thegroup from Li to Cs.

Reason: The effectof increasing size outweighs the increasingnuclear charge, and the outermost electron isvery well screened from the nuclear charge.
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Vishal Baghel

Contributor-Level 10

(d) The ionization enthalpies of the alkali metals are considerably low and decrease down the group from Li to Cs. This is because the effect of increasing size outweighs the increasing nuclear charge, and the outermost electron is very well screened from the nuclear charge.

New question posted

11 months ago

Ncert Solutions Chemistry Class 11th The S Block Elements

Assertion and Reason:

Directions:

(a) Assertion and reason both are correct statements and reason is correct explanation for assertion.

(b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.

(c) Assertion is correct statement but reason is wrong statement.

(d) Assertion is wrong statement but reason is correct statement.

10.1.Assertion: The elements of Group 2 are commonly known asthe alkaline earth metals.

Reason: Theiroxides and hydroxides are alkaline in nature and thesemetal oxides are found in the earth’s crust.
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New question posted

11 months ago

Ncert Solutions Chemistry Class 11th The S Block Elements

Assertion and Reason:

Directions:

(a) Assertion and reason both are correct statements and reason is correct explanation for assertion.

(b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.

(c) Assertion is correct statement but reason is wrong statement.

(d) Assertion is wrong statement but reason is correct statement.

10.1.Assertion: The elements of Group 2 are commonly known asthe alkaline earth metals.

Reason: Theiroxides and hydroxides are alkaline in nature and thesemetal oxides are found in the earth’s crust.
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New answer posted

11 months ago

Ncert Solutions Chemistry Class 11th Some Basic Concepts of Chemistry

1.36. Chlorine is prepared in the laboratory by treating manganese dioxide (MnO2) with aqueous hydrochloric acid according to the reaction.

4 HCl (aq) + MnO2 (s) ———–> 2 H2O (l) + MnCl2(aq) +Cl2(g)

How many grams of HCl react with 5.0 g of manganese dioxide? (Atomic mass of Mn = 55 u
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P
Payal Gupta

Contributor-Level 10

1.36. The molar mass of MnO2is 87g and the molar mass of HCl is 36.5 g.

According to the equation, 4 moles of HCl (i.e. 4 x 36.5g = 146 g) reacts with 1 mol of MnO2 (87g).

So, for 5.0 g of MnO2 will react with:

= 5 x146/87 = 8.40 g of MnO2

Therefore, 8.4 g of HCl will react with 5 g of MnO2.

New answer posted

11 months ago

Ncert Solutions Chemistry Class 11th Some Basic Concepts of Chemistry

1.35. Calcium carbonate reacts with aqueous HCl according to the reaction

CaCO3 (s) + 2HCl (aq) ———->CaCl2 (aq) +CO2(g) +H2O(l).

What mass of CaCO3 is required to react completely with 25 mL of 0.75 M HCl?
0 Follower 10 Views

P
Payal Gupta

Contributor-Level 10

1.35. Step 1: 0.75 M HCl means 0.75 mol per 1000 mL or 0.75 x 36.5 g in 1000 mL.

i.e. 1000 mL of 0.75 M HCl contains 0.75 x 36.5 g HCl

Therefore, 25 mL of 0.75 M HCl contains 0.75 x 36.5 x 25/ 1000 g = 0.6844 g


Step 2: To calculate mass of CaCO3reacting completely with 0.6844 g of HCl
CaCO3  (s) + 2HC1 (aq)———>CaCl2 (aq) +CO2 (g) + H2O
2 mol of HCl, i.e., 2 x 36.5 g = 73 g HCl react completely with CaC03 = 1 mol = 100 g

Therefore, 0.6844 g HCl will react completely with CaCO3 = 100/73 x 0.6844 g = 0.938 g

New answer posted

11 months ago

Ncert Solutions Chemistry Class 11th Some Basic Concepts of Chemistry

1.34. A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gives 3.38 g carbon dioxide, 0.690 g of water and no other products. A volume of 10.0 L (measured at S.T.P.) of this welding gas is found to weigh 11.6 g. Calculate (i) empirical formula, (ii) molar mass of the gas, and (iii) molecular formula.
0 Follower 27 Views

P
Payal Gupta

Contributor-Level 10

1.34.  (i) 1 mole (44 g) of CO2 will have 12 g carbon.

So, 3.38 g of CO2 will have carbon = 12g/44g * 3.38

= 0.9217 g

18 g of water will have 2 g of hydrogen.

So, 0.690 g of water contain hydrogen = 2g/18g * 0.6902g

= 0.0767 g

Since carbon and hydrogen are the only constituents of the compound, the total mass of the compound is:

= 0.9217 g + 0.0767 g

=0.9984 g

So, the percentage of Carbon in the compound = 0.9217/0.9984 * 100 = 92.32%

Now, percentage of Hydrogen in the compound = 0.0767/0.9984 * 100 = 7.68%

Moles of carbon in the compound = 92.32/12=7.69

Moles of hydrogen in the compound = 7.68/1=7.68

Since, we ha

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New answer posted

11 months ago

Ncert Solutions Chemistry Class 11th Some Basic Concepts of Chemistry

1.33. Calculate the number of atoms in each of the following:

(i) 52 moles of He

(ii) 52 u of He

(iii) 52 g of He    
0 Follower 12 Views

P
Payal Gupta

Contributor-Level 10

1.33.  (i) 52 moles of Ar

Ans:1 mole of Ar = 6.022 * 1023 atoms of Ar

Therefore, 52 mole of Ar = 52 * 6.022 * 1023 atoms of Ar= 3.131 * 1025 atoms of Ar

(ii) 52 u of He

Ans:1 atom of He = 4u of the He

Or, 4 u of He = 1 atom of He

So, 52 u of He = 52/4 atom of He = 13 atoms of He.

(iii) 52 g of He

Ans:4g of He = 6.022 * 1023 atoms of He

So, 52g of He = 6.022 * 1023 * 52/4 atoms of He

= 7.8286 * 1024 atoms of He

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