Question

Calculate the mass percentage of benzene (C6H6) and carbon tetrachloride (CCl4) if 22 g of benzene is dissolved in 122 g of carbon tetrachloride.

Answer 1

Hey there!

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Answer :

Given that,

Mass of benzene $(C_{6} H_{6})$ = 22 g

Mass of carbon tetrachloride $(CCl_{4})$ = 122 g

Mass of solution dissolved = 22 g + 122 g = 144 g

Then, we have to find,

Mass percentage of benzene = ?

Mass percentage of carbon tetrachloride = ?

Now, using the formula,

∴ Mass percentage of benzene = $\frac{ \text{Mass of benzene} }{ \text{Mass of benzene + Mass of carbon tetrachloride} }$ × 100

⇒ Mass percentage of benzene = $\frac{22}{22 + 122}$ × 100

⇒ $\frac{22}{144}$ × 100

⇒ $\frac{11}{72}$ × 100

⇒ $\frac{1100}{72}$ = 15.2%

→ Mass percentage of benzene = 15.2%

∴ Mass percentage of $CCl_{4}$ = 100 -15.2% = 84.8%

→ Mass percentage of $CCl_{4}$ = 84.8%

Answer 2

Hey !!

Mass percentage of benzene = Mass of benzene / Mass of solute × 100

           = 22 g / 22 g + 122 g

           = 15.28 %

Mass percentage of carbon tetrachloride = 100 - Mass percentage of benzene

           = 100 - 15.28

FINAL RESULT = 84.72 %

Hope it helps you !!!