1) We know that:

Mass percentage of C6H6

and we also know that:

Mass percentage of CCl4

Alternatively,

Mass percentage of CCl4= (100 - 15.28)%

= 84.72%

2) Let the mass of solution be 100 g

Mass of Benzene = 30 g

Mass of carbon tetrachloride = 100 g - 30 g = 770 g

Number of moles of benzene = Mass / Molar mass

                     = 30 g / 78 g mol⁻¹

                     = 0.385 mol

Number of moles of carbon tetrachloride = Mass / Molar mass

                   = 70 g / 154 g mol⁻¹

                   = 0.455 mol

Mole fraction of benzene = Mass of benzene / Moles of benzene + Moles of tetrachloride

                 = 0.385 mol / 0.840 mol

FINAL RESULT  :- 0.458

3) (a) weight of Co(NO3)2.6H2O = 30g

molar mass of Co(NO3)2.6H2O = 291 g/mol.

volume of solution = 4.3 L

so, number of moles of solute = given weight/molar mass

= 30/291 = 0.103 mol

molarity = 0.103mol/4.3L = 0.023M

(b) Given,

30 mL of 0.5 M H2SO4 diluted to 500 mL.

In 1000 mL of 0.5 M H2SO4, number of moles present is 0.5 mol.

∴ In 30 mL of 0.5 M H2SO4, number of moles present = 30 × 0.5/1000 = 0.015 mol

∴ molarity = number of moles present/volume

= 0.015mol/volume of solution

= 0.015/0.5L

= 0.03M

4) Mass of required aqueous solution = 2.5 kg = 2500 g

0.25 molal aqueous solution of urea it means that 0.25 mole of urea is dissolved in 1000 grams of water.

Mass of water = 1000 g

Mass of urea = 0.25 mol

Molar mass of urea (H₂NCONH₂) = 4 × 1 + 2 × 14 + 1 × 12 + 1 × 16 = 60 g mol⁻¹

        NOTE :- Mass of urea = Number of moles of urea × Molar mass of urea  

     Mass of 0.25 moles of urea = 0.25 mol ×× 60 h mol⁻¹ = 15 g

Mass of solution = 1000 g + 15 g = 1015 g

1015 g of aqueous solution contains urea = 15 g

∴ 2500 g of aqueous solution will require urea  

         = 15 g / 1015 kg × 2500 g  

            = 36.95 g

5) Molar Mass of Kl = 166 g/mol

Density = 1.202 g/ml

Mass/Mass % = 20%

Mass of solution = 100g

Mass of solute = 20g

Mass of solvent = 80g = 0.08 kg

Density = Mass of solution / Volume of solution

1.202 = 100 / V

V = 100/1.202

V = 83.19 ml

V = 0.083 L

Moles of Kl = Given mass/Molar Mass = 20/166 = 0.12

-----------------------------------------------------

a) Molality = Moles of Kl /Mass of solvent (Kg)

= 0.12 / 0.08

= 1.5 m

-----------------------------------------------------

b) Molarity = Moles of solute / Volume of solution

= 0.12 / 0.083

= 1.44 M

------------------------------------------------------

c) Mole Fraction = Mole of solute / Moles of solute + Moles of solvent

= 0.12 / ( 0.12 + 4.44 )

= 0.12 / 4.56

= 0.026

-------------------------------------------------------

HOPE THIS HELPS....

@HARSHA