Football, also called association football or soccer, is a game involving two teams of 11 players who try to maneuver the ball into the other team’s goal without using their hands or arms. The team that scores more goals wins. Football is the world’s most popular ball game in numbers of participants and spectators.
Answer:
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Football, also called association football or soccer, is a game involving two teams of 11 players who try to maneuver the ball into the other team’s goal without using their hands or arms. The team that scores more goals wins. Football is the world’s most popular ball game in numbers of participants and spectators.
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Answer:
Mass percentage of C6H6
Mass percentage of CCl4
Alternatively,
Mass percentage of CCl4 = (100 − 15.28)%
= 84.72%
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Question 2:
Calculate the mole fraction of benzene in solution containing 30% by mass in carbon tetrachloride.
Solution 2:
Let the total mass of the solution be 100 g and the mass of benzene be 30 g.
∴Mass of carbon tetrachloride = (100 − 30)g
= 70 g
Molar mass of benzene (C6H6) = (6 × 12 + 6 × 1) g mol−1
= 78 g mol−1
∴Number of moles of
= 0.3846 mol
Molar mass of carbon tetrachloride (CCl4) = 1 × 12 + 4 × 35.5
= 154 g mol−1
∴Number of moles of CCl4
= 0.4545 mol
Thus, the mole fraction of C6H6 is given as:
= 0.458
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Question 3:
Calculate the molarity of each of the following solutions: (a) 30 g of Co(NO3)2. 6H2O in 4.3 L of solution (b) 30 mL of 0.5 M H2SO4 diluted to 500 mL.
Solution 3:
Molarity is given by:
(a) Molar mass of Co (NO3)2.6H2O = 59 + 2 (14 + 3 × 16) + 6 × 18
= 291 g mol−1
∴Moles of Co (NO3)2.6H2O
= 0.103 mol
Therefore, molarity
= 0.023 M
(b) Number of moles present in 1000 mL of 0.5 M H2SO4 = 0.5 mol
∴Number of moles present in 30 mL of 0.5 M H2SO4
= 0.015 mol
Therefore, molarity
= 0.03 M
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Question 4:
Calculate the mass of urea (NH2CONH2) required in making 2.5 kg of 0.25 molal aqueous solution.
Solution 4:
Molar mass of urea (NH2CONH2) = 2(1 × 14 + 2 × 1) + 1 × 12 + 1 × 16
= 60 g mol−1
0.25 molar aqueous solution of urea means:
1000 g of water contains 0.25 mol = (0.25 × 60)g of urea
= 15 g of urea
That is,
(1000 + 15) g of solution contains 15 g of urea
Therefore, 2.5 kg (2500 g) of solution contains
= 36.95 g
= 37 g of urea (approximately)
Hence, mass of urea required = 37 g
Note: There is a slight variation in this answer and the one given in the NCERT textbook.
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Question 5:
Calculate (a) molality (b) molarity and (c) mole fraction of KI if the density of 20% (mass/mass) aqueous KI is 1.202 g mL-1.
Solution 5:
(a) Molar mass of KI = 39 + 127 = 166 g mol−1
20% (mass/mass) aqueous solution of KI means 20 g of KI is present in 100 g of solution.
That is,
20 g of KI is present in (100 − 20) g of water = 80 g of water
Therefore, molality of the solution
= 1.506 m
= 1.51 m (approximately)
(b) It is given that the density of the solution = 1.202 g mL−1
∴Volume of 100 g solution
= 83.19 mL
= 83.19 × 10−3 L
Therefore, molarity of the solution
= 1.45 M
(c) Moles of KI
Moles of water
Therefore, mole fraction of KI
= 0.0263
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All Exercises – Chapter 2 – Solutions
Intext Exercise 2.1
Intext Exercise 2.2
Intext Exercise 2.3
Intext Exercise 2.4
NCERT Exercise 2
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