Question

A solution of benzene in methanol has a transmittance of 28% in a 1.00 cm cell at a wavelength of 254 nm. only the benzene absorbs light at this wavelength, not the methanol. what will the solution's transmittance become (in the same 1.00 cm cell) if the original benzene solution is diluted 10-fold with methanol?

Answer 1

Answer:

$\boxed{88 \%}$

Explanation:

$A = \epsilon cl\\A = -\log T\\\epsilon cl = -\log T$

ε and l are constants.

If there are two different concentrations of benzene,

$\dfrac{c_{2}}{c_{1}} = \dfrac{\log T_{2}}{\log T_{1}}$

Data:

c₁ = c₁; c₂ = 0.1c₁;

T₁ = 0.28: T₂ = ?

Calculation:

$\dfrac{0.10c_{1}}{c_{1}} = \dfrac{\log T_{2}}{\log 0.28}\\\\\log T_{2}= 0.10 \times \log0.28 = -0.0553\\\\T_{2} = 10^{-0.0553} = 0.88 \\\\\boxed{T_{2} = \textbf{88 \%}}$