11th Std Chemistry Guide PDF Free Download

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Question 1.
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
(a) 40 ml CO₂
(b) 40 ml CO₂ gas and 80 ml H₂o gas
(c) 60 ml CO₂ gas and 60 ml H₂o gas
(d) 120 ml CO₂ gas
Answer:
(a) 40 ml CO₂
Solution:
CH₄(g) + 2O₂ → CO₂(g) + 2H₂O (1)

Since the product was cooled to room temperature, water exists mostly as liquid. Hence, option (a) is correct.

Question 2.
An element X has the following isotopic composition ²⁰⁰X = 90%, ¹⁹⁹X = 8% and ²⁰²X = 2%. The weighted average atomic mass of the element X is closest to
(a) 201 u
(b) 202 u
(c) 199 u
(d) 200 u
Answer:
(d) 200 u
Solution:
X = (200×90)+(199×8)+(202×2)100 = 199.96 = 200 u

Question 3.
Assertion:
Two mole of glucose contains 12.044 × 10²³ molecules of glucose
Reason:
Total number of entities present in one mole of any substance is equal to 6.02 × 10²²
(a) both assertion and reason are true and the reason is the correct explanation of assertion
(b) both assertion and reason are true but reason is not the correct explanation of assertion
(c) assertion is true but reason is false
(d) both assertion and reason are false
Answer:
(c) assertion is true but reason is false
Solution:
Correct reason:
Total number of entities present in one mole of any substance is equal to 6.022 × 10²³.

Question 4.
Carbon forms two oxides, namely carbon monoxide and carbon dioxide. The equivalent mass of which element remains constant?
(a) Carbon
(b) Oxygen
(c) both carbon and oxygen
(d) neither carbon nor oxygen
Answer:
(b) Oxygen
Solution:
Reaction 1:
2C + O₂ → 2CO
2 × 12g carbon combines with 32g of oxygen. Hence, Equivalent mass of carbon
2×1232 × 8 = 6
Reaction 2:
C + O₂ → CO₂
12 g carbon combines with 32 g of oxygen. Hence, Equivalent mass of carbon
= 1232 × 8 = 6