Question

At a certain wavelength, liquid P transmits 70%, whereas liquid Q transmits 30% of the incident light when separately placed in a spectrophotometric cell (path length = 1 cm). In a binary mixture of liquids P and Q (assume non-interacting liquids), the absorbance in the same cell is 0.25. The volume fraction of liquid P in the binary mixture is .......... (Round off to two decimal places)

Hint:

The Beer-Lambert law can be used to calculate the absorbance of a mixture by adding the absorbances of individual components based on their transmission.

Answer 1

Correct Answer: 0.73 - 0.75

To solve this problem, we need to find the volume fraction of liquid P in the binary mixture using absorbance and transmission data. The relationship between transmission (T) and absorbance (A) is \( A = -\log(T) \). Let's analyze the given data:

1. Transmission of liquid P (\(T_P\)) = 70% = 0.70 results in absorbance \(A_P = -\log(0.70)\).
2. Transmission of liquid Q (\(T_Q\)) = 30% = 0.30 results in absorbance \(A_Q = -\log(0.30)\).

First, calculate the absorbance values:
\(A_P = -\log(0.70) \approx 0.155\)
\(A_Q = -\log(0.30) \approx 0.523\)

Let \(x\) be the volume fraction of liquid P, and hence \(1-x\) is the volume fraction of liquid Q. In a binary mixture, absorbance is additive:
\(A_{\text{mixture}} = x \cdot A_P + (1-x) \cdot A_Q\)

Given: \(A_{\text{mixture}} = 0.25\)
Substituting the values:
\(0.25 = x \cdot 0.155 + (1-x) \cdot 0.523\)
\(0.25 = 0.155x + 0.523 - 0.523x\)
Solving for \(x\):
\(0.25 = 0.523 - 0.368x\)
\(0.368x = 0.523 - 0.25\)
\(0.368x = 0.273\)
\(x = \frac{0.273}{0.368}\)
\(x \approx 0.74\)

Thus, the volume fraction of liquid P in the binary mixture is \(0.74\).

Therefore, the volume fraction of liquid P is 0.74.