A solution containing 47 ppm of a compound X (FW 225) has a transmittance of $29.7 \%$ in a 1.5 cm cell at 400 nm. The molar absorptivity in $\mathrm{L} \mathrm{mol}^{-1} \mathrm{cm}^{-1}$ is
(a)$1.89 \times 10^{3}$
(b)$1.42 \times 10^{3}$
(c)$1.68 \times 10^{3}$
(d)$1.79 \times 10^{3}$
Answer
Answer: C ✓ checked by 4AB · confidence high
Explanation
A = −log 0.297 = 0.527; c = 0.047 g L⁻¹/225 = 2.09×10⁻⁴ M; ε = 0.527/(1.5 × 2.09×10⁻⁴) ≈ 1.68×10³ (c).