Question

calculate the molality and mole fraction of 2.5 gm of acetic acid in 75 gm of c6h6

Answer 1

Answer:

molality =?

w of CH3COOH=2.5 gm

mol.wt. of CH3COOH= 12+3+12+16 x2 +1=60

w of benzene= 75 gm =0.075 kg

mol.wt. of C6H6= 12 x6 +1 x 6 =78

Molality =wt. of solute/ (mol.wt. of solute x wt. of solvent in kg)

m=2.5/60 x 0.075

m =0.56 mole /kg Ans.

(B) mole fraction

moles of CH3COOH= wt. /mol.wt =2.5/60

moles of CH3COOH=0.04

moles of C6H6= wt. /mol.wt =75/78

moles of C6H6=0.96

total moles =0.042 +0.96=1.0

mole fraction of CH3COOH = moles of CH3COOH/total moles=0.04/1 =0.04

mole fraction of C6H6 = moles of C6H6 /total moles=0.96/1 =0.96

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1.Mole fraction is defined as= Number of moles of solute/ Number of moles of solution.

Molecular mass of ethanoic acid=60

Molecular mass of benzene=78

Number of moles of ethanoic acid= 2.5/60=0.041

Number of moles of benzene= 75/78=0.96153

Mole fraction= 0.041/(0.041+0.96153)=0.040

2. Molality = Number of moles of solute/weight of the solvent (in kg)

= 0.041/0.075

=0.5466