I want the solution of Q.3,Q.4 for the questions you can the attachment?
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3)
Gondia1
Gondia1
29.09.2020
Chemistry
Secondary School
answered
when 1.80g of a non volatile compound are dissolved in 25.0g of acetone, the solution boils at 56.86°C while pure acetone boils at 56.38°C under the same atmospheric pressure. calculate the molecular weight of the compound. The molal elevation constant for acetone is 1.72 K Kg mol-1
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abhi178
abhi178
Given info : when 1.80g of a non volatile compound are dissolved in 25.0g of acetone, the solution boils at 56.86°C while pure acetone boils at 56.38°C under the same atmospheric pressure. The molal elevation constant for acetone is 1.72 K kg/mol
To find : The molecular weight of the compound is...
solution : molality, m = no of moles of solute/mass of solvent in kg
= (1.8/M)/(25/1000)
= 72/M
where M is molecular weight of solute.
now using formula, ∆Tb = K_b × m
here K_b is molal elevation constant, m is molality and ∆T_b is elevation of boiling point.
∆T_b = 56.86°C - 56.38°C = 0.48°C
m = 72/M
K_b = 1.72 K kg/mol
∴ 0.48 = 1.72 × 72/M
⇒M = 72 × 1.72/0.48 = 72 × 4 = 288 g/mol
Therefore the molecular weight of the compound is 288 g/mol.
4) given, mass of Urea = 5g molar mass of urea = 60g/mol so, number of mole of urea = 5/60=1/12 mass of solvent (alcohol) = 75g we know, molality = number of mole of solute/mass of solvent in kg = (1/12) 1000/75 = = 1000/(12 x 75) = 1.11molal....(1) now using formula, ATb = Kb x m where ATB is elevation in boiling point, Kb is elevation constant and m is molality. given, Kb = 1.15 °C/molal, m =1.11 molal so, ATb = 1.15 x 1.11 1.2765°C
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