Ncert Solutions Chemistry Class 11th
Get insights from 2k questions on Ncert Solutions Chemistry Class 11th, answered by students, alumni, and experts. You may also ask and answer any question you like about Ncert Solutions Chemistry Class 11th
- 2k
Questions
- 0
Discussions
- 0
Active Users
- 29
Followers
Have a question about Ncert Solutions Chemistry Class 11th? Ask our experts
Keep it short & simple. Type complete word. Avoid abusive language.
10.4. Assertion: Alkali metals are always univalent.
Reason: Alkali metals are soft.
10.4. Assertion: Alkali metals are always univalent.
Reason: Alkali metals are soft.
(b) Both the statements are correct but not the reason for the assertion.
Upvote 
<p><strong>10.3. Assertion: The alkalimetals and their salts impart characteristiccolour to an oxidizing flame. </strong></p>
<p><strong>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.</strong></p>
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.
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.
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.
Upvote 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.
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.
(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.
Upvote <p><strong>Assertion and Reason:</strong></p>
<p>Directions:</p>
<p>(a) Assertion and reason both are correct statements and reason is correct explanation for assertion.</p>
<p>(b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.</p>
<p>(c) Assertion is correct statement but reason is wrong statement.</p>
<p>(d) Assertion is wrong statement but reason is correct statement.</p>
<p><strong>10.1.Assertion: The elements of Group 2 are commonly known asthe alkaline earth metals. </strong></p>
<p><strong>Reason: Theiroxides and hydroxides are alkaline in nature and thesemetal oxides are found in the earth’s crust.</strong></p>
<p><strong>Assertion and Reason:</strong></p>
<p>Directions:</p>
<p>(a) Assertion and reason both are correct statements and reason is correct explanation for assertion.</p>
<p>(b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.</p>
<p>(c) Assertion is correct statement but reason is wrong statement.</p>
<p>(d) Assertion is wrong statement but reason is correct statement.</p>
<p><strong>10.1.Assertion: The elements of Group 2 are commonly known asthe alkaline earth metals. </strong></p>
<p><strong>Reason: Theiroxides and hydroxides are alkaline in nature and thesemetal oxides are found in the earth’s crust.</strong></p>
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
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
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.
Upvote 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?
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?
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
Ca
Upvote 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.
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.
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 compo
Upvote 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
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
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
Upvote