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# Calculate the mass percentage of benzene C_6 H_6 and carbon tetrachloride CCl_4 if 22 g of benzene is dissolved in 122 g of carbon tetrachloride.

2.1   Calculate the mass percentage of benzene $\inline (C_{6}H_{6})$  and carbon tetrachloride $\inline (CCl_{4})$  if $\inline 22\; g$ of benzene is dissolved in $\inline 122\; g$ of carbon tetrachloride.

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We know that solute and solvent forms solution.

So mass percentage of benzene (solute)   :-

$=\frac{22}{22+122}\times100 = \frac{22}{144}\times100 = 15.28\%$

Similarly mass percentage of CCl4 :-

$=\frac{122}{22+122}\times100 = \frac{122}{144}\times100 = 84.72\%$

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