Answer:
[tex]\boxed{88 \%}[/tex]
Explanation:
[tex]A = \epsilon cl\\A = -\log T\\\epsilon cl = -\log T[/tex]
ε and l are constants.
If there are two different concentrations of benzene,
[tex]\dfrac{c_{2}}{c_{1}} = \dfrac{\log T_{2}}{\log T_{1}}[/tex]
Data:
c₁ = c₁; c₂ = 0.1c₁;
T₁ = 0.28: T₂ = ?
Calculation:
[tex]\dfrac{0.10c_{1}}{c_{1}} = \dfrac{\log T_{2}}{\log 0.28}\\\\\log T_{2}= 0.10 \times \log0.28 = -0.0553\\\\T_{2} = 10^{-0.0553} = 0.88 \\\\\boxed{T_{2} = \textbf{88 \%}}[/tex]