Chemistry Class Twelve

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Solutions

Solution of NCERT In Text Questions

Question - 2.1 - Calculate the mass percentage of benzene (C6 H 6) and carbon tetrachloride (CCl 4) if 22 g of benzene is dissolved in 122 g of carbon tetrachloride.

Solution:

Given,

Mass of solute (WB ) = 22g

Mass of solvent (W B) = 122 g

Mass of solution = 22g + 122 g = 144 g

Mass % of benzene = ?

Mass % of carbon tetrachloride = ?

Now, Mass percentage of benzene

Solutions class 12 chemistry - NCERT In Text Solution2

And, Mass percentage of carbon tetrachloride Solutions class 12 chemistry - NCERT In Text Solution3

Solutions class 12 chemistry - NCERT In Text Solution4

Thus, Mass percentage of benzene = 15.27%

And, Mass percentage of carbon tetrachloride = 84.72%

Question – 2.2 - Calculate the mole fraction of benzene in solution containing 30% by mass in carbon tetrachloride.

Solution:

Given, % of benzene by mass = 30%

Thus, % of solvent (carbon tetrachloride) = 70%

Mole fraction of benzene in the given solution = ?

This means, 30g of benzene (C6 H 6) is dissolved in 70 g of CCl 4

Now, Molecular mass of benzene (C6 H6 ) = (12 x 6) + (1 x 6) = 78 g mol – 1

Molecular mass of carbon tetrachloride (CCl4 ) = 12 + (35.5 x 4 ) = 154 g mol – 1

Therefore, Number of moles of C 6H 6 Solutions class 12 chemistry - NCERT In Text Solution5

Solutions class 12 chemistry - NCERT In Text Solution6

And, number of moles of CCl4 Solutions class 12 chemistry - NCERT In Text Solution7

Solutions class 12 chemistry - NCERT In Text Solution8

Thus, mole fraction of benzene in the given solution

Solutions class 12 chemistry - NCERT In Text Solution9
Solutions class 12 chemistry - NCERT In Text Solution10

Thus, mole fraction of benzene in the given solution = 0.457

Question – 2.3 - Calculate the molarity of each of the following solutions:

(a) 30 g of Co(NO 3) 2. 6H2 O in 4.3 L of solution

(b) 30 mL of 0.5 M H 2SO4 diluted to 500 mL.

Solution:

(a) Given,

Mass of solute (W B) = 30 g

Molar mass of given solute Co(NO 3) 2.6H 2O = 58.7 + 2[14 + (16 x 3)] + 6 ( 2 + 16)

= 58.7 + (2 x 62) + (6 x 18)

= 58.7 + 124 + 128 g mol – 1

= 290.7 g mol – 1

Now, Number of moles of Co(NO 3)2 .6H 2O Solutions class 12 chemistry - NCERT In Text Solution11

Solutions class 12 chemistry - NCERT In Text Solution12

Now, we know that, Molarity Solutions class 12 chemistry - NCERT In Text Solution13

Solutions class 12 chemistry - NCERT In Text Solution14

Thus, molarity of given solute = 0.24 M

(b) Given, 30 mL of 0.5 M H2SO4 diluted to 500 mL.

Thus,

Solutions class 12 chemistry - NCERT In Text Solution15

Thus, required molarity = 0.3 M

Alternate method:

Number of moles present in 1000 ml of 0.5 M H 2SO4 solution = 0.5 mol

Therefore, number of moles present in 1 ml of solution = 1 / 1000 mL

Therefore, number of moles present in 30 mL of solution Solutions class 12 chemistry - NCERT In Text Solution16

Now, we know that, Molarity Solutions class 12 chemistry - NCERT In Text Solution17

Solutions class 12 chemistry - NCERT In Text Solution18

Question – 2.4 - Calculate the mass of urea (NH2 CONH 2) required in making 2.5 kg of 0.25 molal aqueous solution.

Solution: Given,

Molality (m) = 0.25 m

Molar mass (M B) of urea (NH2 CONH 2) = 14 + 2 + 12 + 16 + 14 + 2 = 60 g mol – 1

Mass of solvent (WA ) = 2.5 kg = 2500 gm

Mass of solute (W B) = ?

Now, we know that,

Solutions class 12 chemistry - NCERT In Text Solution19
Solutions class 12 chemistry - NCERT In Text Solution20

Thus, mass of given solute urea = 37.50 g

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Solutions - Class 12 chemisty - NCERT In Text Solution (Part 2)
Solutions - Class 12 chemisty - NCERT In Text Solution (Part 3)


 
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