Chemistry

10.5. Assertion: Electrode potential is a measure of the tendency of an element to lose electrons in the aqueous solution. So, the reducing property can be correlated in terms ofelectrode potentials (E°) of alkali metals.

Reason: More positive is the electrode potential, higher is the tendency of the element to loose electrons and hence, stronger is the reducing agent.

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Vishal Baghel

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(c) Reason is wrong a statement. More negative is the electrode potential, higher is the tendency of the element to lose electrons and hence, stronger is the reducing agent.

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10.4. Assertion: Alkali metals are always univalent.

Reason: Alkali metals are soft.

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Vishal Baghel

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

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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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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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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.

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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.

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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.

(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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a year ago

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.

(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

a year ago

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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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.

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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?

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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

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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.

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