Question

calculate the mole fraction of 4g of ethanoic acid (CH3COOH)in 80g of benzene (c6h6)​

Answer 1

Answer:

13/200

Explanation:

No of moles in 4g of ethanoic acid = 4/60 = 1/15

No of moles in 80g of Benzene = 80/78 = 40/39

Mole fraction = 1/15 ÷ 40/39

= 1/15 × 39/40 = 13/200

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Answer 2

Answer:

Mole fraction = 13/200

Explanation:

CH3COOH - Molar mass = 60 g

Moles present in 4g CH3COOH = 4/60

C6H6 Molar mass = 78 g

Moles present in 80g C6H6 = 80/78

so the mole fraction is = 1/15 ÷ 80/78

                                      = 1/15 × 78/80

                                      = 1/15 × 39/40

                                      = 39/600

                                      = 13/200