Question

Calculate the mole fraction of benzene in solution containing 30% by mass in carbon tetrachloride.

Answer 1

Benzene = C6H6

Carbon Tetrachloride = CCl4

Molar Mass of C6H6 = 78 g/mol

Molar mass of CCl4 = 154 g/mol

Mass/Mass % = 30%

Mass of solution = 100g

Mass of solute ( C6H6 ) = 30g

Mass of solvent ( CCl4 ) = 70g

Mole of C6H6 = Given mass/Molar mass
= 30/78
= 0.38

Moles of CCl4 = Given mass/Molar mass
= 70/ 154
= 0.45


Mole Fraction = Moles of C6H6 / ( Moles of C6H6 + Moles of CCl4 )

= 0.38 / ( 0.38 + 0.45 )

= 0.38 / 0.83

= 0.45

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Thanks !

Answer 2

Hey there!

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Answer :

Given,

Mass/Mass percentage = 30 %

So, lets take mass of the solution = 100g

Mass of benzene $(C_{6} H_{6})$ = 30% of 100g = 30g

Mass of carbon tetrachloride $(CCl_{4})$ = total mass - mass of benzene

∴ (100 − 30)g = 70 g

→ Molar mass of benzene $(C_{6} H_{6})$ = 6 × 12 + 6 × 1 = 78 g $mol^{-1}$

→ Molar mass of carbon tetrachloride $CCl^{4}$ = 1 × 12 + 4 × 355 = 154 g $mol^{-1}$

∴ $X_{Benzene} = \frac{X_{Benzene}}{X_{Benzene} + n_{ccl_{4}}}$

∴ $X_{Benzene} = \frac{\frac{W_{Benzene}}{M_{Benzene}}}{\frac{W_{Benzene}}{M_{Benzene} } + \frac{W_{CCl_{4}}}{M_{CCl_{4}}}}$

∴ $X_{Benzene} = \frac{\frac{30}{78}}{\frac{30}{78}+ \frac{70}{154}}$

∴ $X_{Benzene} = \frac{\frac{30}{78}}{\frac{30 × 154 + 78 × 70}{154 × 78}}$

∴ $X_{Benzene} = \frac{30 × 154}{4620 + 5460} = \frac{4620}{10080}$

⇒ $X_{Benzene}$ = 0.458