(6 ~g ) of a mixture of naphthalene ( ( C ₁₀ H ₈ ) ) and anthracene ( ( C ₁₄ H ₁₀ ) ) is dissolved in 300 gram of benzene. If the depression in freezing point is (0.70 ~K ), the composition of naphthalene and anthracene in the mixture respectively in ( g ) are (molal depression constant of benzene is (5.1 ~K ~kg ~mol ⁻¹ ) )

Step-by-step solution

Given, weight of solvent (=300 ~g ) ( aligned Molality & = moles of solute weight of solvent ( kg ) K_f & =5.1 ~K ~kg / mol aligned ) Depression in freezing point, ( gathered T_f=K_f m 0.70=51 m 0.70 5.1 = Total moles of solute Total weight of solvent ( kg ) gathered ) Lets, assume (x ) g napthalene is present, ( aligned 0.137 & = x w_ C ₁₀ H ₈ + 6-x w_ C ₁₄ H ₁₀ 1000 300 0.041 & = x 128 ~g + 6-x 178 ~g 25 x=84.211 x( napthalene ) & =3.4 Now, anthracene & =6-x=6-3.4=2.6 ~g aligned ) Hence, option (2) is correct.