Two elements A and B form compounds having formula `AB_2` and `AB_4` . When dissolved in 20 g of benzene `(C_6H_6)`, 1 g of `AB_2` lowers the freezing point by 2.3 K whereas 1.0 g of `AB_4` lowers it by 1.3 K. The molar depression constant for benzene is 5.1 K kg `"mol"^(-1)` . Calculate atomic masses of A and B.

Explanation

Using the formula, `M_2 = (1000K_f w_2)/(w_1 xx Delta T_f)`
Molar mass of `AB_2 , M_(AB_2) = (1000xx 5.1 xx 1)/(20 xx 2.3) = 110.87 g "mol"^(-1)`
Molar mass of `AB_4 , M_(AB_4) = (1000 xx 5.1 xx 1)/(20 xx 1.3) = 19615 g "mol"^(-1)`
Atomic mass of A + 2 (Atomic mass of B) = 110.87 ....(i)
Atomic mass of A + 4 (Atomic mass of B) = 196.15 ...(ii)
Equation (ii) - Equation (i) gives 2 (Atomic mass of B) = 85.28 or Atomic mass of B = 42.64 u
Substituting in equation (i), we get
Atomic mass of A + 85.28 = 110.87
Atomic mass of A = 25.59 u.