Question

calculate molality of 2.5 gram of ethanoic acid in 80 g of benzene​

Answer 1

Answer:

The molar mass of ethanoic acid is 2(12)+2(16)+4(1)=60 g/mol. The mass of ethanoic acid is divided with molar mass of ethanoic acid to obtain number of moles of ethanoic acid.

Number of moles of ethanoic acid =

60g/mol

2.5g

​

=0.04167mol

The mass of benzene is 80 grams or

1000/80

​

=0.080 kg.

The molality of ethanoic acid in benzene is the ratio of the number of moles of ethanoic acid to the mass of benzene (in kg).

It is

0.080kg/0.04167mol

​

=0.520875m

Explanation:

Hope it's helpful

Answer 2

Formula of ethanoic acid- CH3COOH

R.M.M= 12+ 3+12+2*16+1

=>28+32=> 60 g mol-1

so 2.5 g of Ethanoic acid is x mol.

2.5/60=x

25/600=x

1/24=x

x=1/24 mol

80 g of benzene= 80/1000 kg =>2/25 kg

molality=mols of solute/ kgs of solvent

(1/24)/(2/25)

25/48

0.52 mol kg-1 is molality of the solution.