1.22 g C6H5COOH is added into the 100 g solvent CH3COCH3 where data of Δ Tb and Kb are given as: ΔTb = 0.17, solute by Kb = 1.7 K kg/mol Find out the molecular weight (g/mol) of the solute in the solution by interpreting the result.

Question

1.22 g C6H5COOH is added into the 100 g solvent CH3COCH3 where data of Δ Tb and Kb are given as: ΔTb = 0.17, solute by Kb = 1.7 K kg/mol Find out the molecular weight (g/mol) of the solute in the solution by interpreting the result. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Δ textT textb = text0.17 = textK textb ⋅ textm text0.17 = text1.7 × ( text1.22/M/ text100/1000) M = 122 g/mole Benzoic acid is not associated or dissociated.

Q. 1.22 g C₆H₅COOH is added into the 100 g solvent CH₃COCH₃ where data of Δ T_band K_b are given as:
ΔT_b = 0.17, solute by K_b = 1.7 K kg/mol
Find out the molecular weight (g/mol) of the solute in the solution by interpreting the result.

$Δ T_{b} = 0.17 = K_{b} \cdot m$
$0.17 = 1.7 \times \frac{1.22/M}{100/1000}$
M = 122 g/mole
Benzoic acid is not associated or dissociated.

Q. 1.22 g C6H5COOH is added into the 100 g solvent CH3COCH3 where data of Δ Tb and Kb are given as: ΔTb = 0.17, solute by Kb = 1.7 K kg/mol Find out the molecular weight (g/mol) of the solute in the solution by interpreting the result.

ΔTb​=0.17=Kb​⋅m 0.17=1.7×100/10001.22/M​ M = 122 g/mole Benzoic acid is not associated or dissociated.

Q. 1.22 g C₆H₅COOH is added into the 100 g solvent CH₃COCH₃ where data of Δ T_band K_b are given as:
ΔT_b = 0.17, solute by K_b = 1.7 K kg/mol
Find out the molecular weight (g/mol) of the solute in the solution by interpreting the result.

$Δ T_{b} = 0.17 = K_{b} \cdot m$
$0.17 = 1.7 \times \frac{1.22/M}{100/1000}$
M = 122 g/mole
Benzoic acid is not associated or dissociated.