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11 months ago

Chemistry 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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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.

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11 months ago

Chemistry 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.

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11 months ago

Chemistry 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.

(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

Chemistry 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.

(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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11 months ago

Chemistry 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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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.

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11 months ago

Chemistry 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?

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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

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11 months ago

Chemistry 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.

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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

...more

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11 months ago

Chemistry 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

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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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11 months ago

Chemistry Some Basic Concepts of Chemistry

1.32. Use the data given in the following table to calculate the molar mass of naturally occurring argon isotopes:

Isotope	Isotopic molar mass	Abundance
³⁶Ar	35.96755g mol^-1	0.337%
³⁸Ar	37.96272 g mol^-1	0.063%
⁴⁰Ar	39.9624 g mol^-1	99.600%

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Payal Gupta

Contributor-Level 10

1.32. Molar mass of argon is

= [ (35.96755 * 0.337/100)+ (37.96272 * 0.063/100)+ (39.9624 * 99.60/100)]g mol^-l

= [0.121+0.024+39.802] g mol^-l

= 39.947 g mol^-l

So, the molar mass of argon is 39.947 g/ mol.

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11 months ago

Chemistry Some Basic Concepts of Chemistry

1.31. How many significant figures should be present in the answer of the following calculations?

(i)

(ii) 5 * 5.364

(iii) 0.0125 + 0.7864 + 0.0215

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Payal Gupta

Contributor-Level 10

1.31. First we need to find the least precise number to find the significant figures.

(i) Least precise number of the calculation is 0.112

Therefore, number of significant figures in the answer = Number of significant figures in the least precise number, i.e. 3

(ii) Least precise number of calculations = 5.364

Therefore, number of significant figures in the answer will be = Number of significant figures in 5.364 = 4

(iii) 0.0125+0.7864+0.0215

Since the least number of decimal places in each term is four, the number of significant figures in the answer will also be 4.

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