Chemistry, asked by meethoney05, 10 months ago

a solution containing 2.2 grams of a substance in 125 grams of benzene boils at 80. 66 degree Celsius kb for 1 kg of benzene is 33. 28 degree Celsius/ molal, the molecular weight of the substance shall be

Answers

Answered by Alleei
2

Answer : The molecular weight of the substance will be, 130.7 g/mol

Explanation :

Formula used for Elevation in boiling point :

\Delta T_b=i\times k_b\times m

or,

T_b-T^o_b=i\times k_b\times \frac{w_2\times 1000}{M_2\times w_1}

where,

T_b = boiling point of solution = 80.66^oC

T^o_b = boiling point of benzene = 80.1^oC

k_b = boiling point constant  of benzene for 1 kg = 33.28^oC/m

Boiling point constant  of benzene for 125 g = \frac{125}{1000}\times 33.28=4.16^oC/m

m = molality

i = Van't Hoff factor = 1 (for non-electrolyte)

w_2 = mass of solute = 2.2 g

w_1 = mass of solvent (benzene) = 125 g

M_2 = molar mass of solute = ?

Now put all the given values in the above formula, we get:

(80.66-80.1)^oC=1\times (4.16^oC/m)\times \frac{(2.2g)\times 1000}{M_2\times (125g)}

M_2=130.7g/mol

Therefore, the molecular weight of the substance will be, 130.7 g/mol

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