Question

The transmittance of a particular solution measured is $T$. The concentration of the solution is now doubled. Assuming that Beer-Lambert’s law holds good for both the cases, the transmittance for the second would be:

• A. $\dfrac{T}{2}$
• B. $2T$
• C. $T^2$
• D. $\sqrt{T}$

Hint:

In Beer-Lambert’s law, doubling concentration doubles absorbance and squares the transmittance: $T' = T^2$.

Answer 1

The Correct Option is C

According to Beer-Lambert's law, the absorbance $A$ is proportional to the concentration $C$:
$A = \log_{10}\left(\dfrac{1}{T}\right) = \varepsilon C l$,
where $T$ is the transmittance, $\varepsilon$ is the molar absorptivity, and $l$ is the path length.
If the concentration is doubled, i.e., $C' = 2C$, then:
$A' = \varepsilon \cdot 2C \cdot l = 2A$
Now from Beer’s law again:
$A' = \log_{10}\left(\dfrac{1}{T'}\right) = 2 \log_{10}\left(\dfrac{1}{T}\right)$
So,
$\log_{10}\left(\dfrac{1}{T'}\right) = \log_{10}\left(\dfrac{1}{T^2}\right)$
$⇒ T' = T^2$
Hence, the new transmittance is $T^2$.