Find out the molecular weight of C6H5COOH in the case where, 1.22gC6H5COOH is added into the solvent benzene and interpret the result. In 100g benzene, ΔTb=0.13 and kb=2.6Kg/mol .

Question

Find out the molecular weight of C6H5COOH in the case where, 1.22gC6H5COOH is added into the solvent benzene and interpret the result. In 100g benzene, ΔTb=0.13 and kb=2.6Kg/mol . (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Δ textT b = text0.13 = ( text2.6) ( text1.22/M/( text100) text1000) text0.13 = text2.6 × ( text1.22/ textM) text10 textM = text2.6 × ( text1.22 × text10/ text0.13) = text244 g/mole This value of molar mass suggest that benzoic acid(C6H5 COOH) is dimerised in benzene

Q. Find out the molecular weight of $C_{6} H_{5} COO H$ in the case where, $1.22 g C_{6} H_{5} COO H$ is added into the solvent benzene and interpret the result.
In $100 g$ benzene, $Δ T_{b} = 0.13$ and $k_{b} = 2.6 K g / m o l$ .

$Δ T b = 0.13 = (2.6) \frac{1.22/M}{\frac{100}{1000}}$
$0.13 = 2.6 \times \frac{1.22}{M} 10$
$M = 2.6 \times \frac{1.22 \times 10}{0.13}$
$= 244 g/mole$
This value of molar mass suggest that benzoic acid(C₆H₅ COOH) is dimerised in benzene

Q. Find out the molecular weight of C6​H5​COOH in the case where, 1.22gC6​H5​COOH is added into the solvent benzene and interpret the result. In 100g benzene, ΔTb​=0.13 and kb​=2.6Kg/mol .

ΔTb=0.13=(2.6)1000100​1.22/M​ 0.13=2.6×M1.22​10 M=2.6×0.131.22×10​ =244 g/mole This value of molar mass suggest that benzoic acid(C6H5 COOH) is dimerised in benzene

Q. Find out the molecular weight of $C_{6} H_{5} COO H$ in the case where, $1.22 g C_{6} H_{5} COO H$ is added into the solvent benzene and interpret the result.
In $100 g$ benzene, $Δ T_{b} = 0.13$ and $k_{b} = 2.6 K g / m o l$ .

$Δ T b = 0.13 = (2.6) \frac{1.22/M}{\frac{100}{1000}}$
$0.13 = 2.6 \times \frac{1.22}{M} 10$
$M = 2.6 \times \frac{1.22 \times 10}{0.13}$
$= 244 g/mole$
This value of molar mass suggest that benzoic acid(C₆H₅ COOH) is dimerised in benzene