Learn Intext Exercise 2.1 with Free Lessons & Tips

Mass percentage of C₆H₆

Mass percentage of CCl₄

Alternatively,

Mass percentage of CCl₄ = (100 − 15.28)%

= 84.72%

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

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

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.