# 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.

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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