A solution containing 3.75 mg/100 ml of X (335 g/mol) has a transmittance of 39.6% in a 1.50-cm cell at 425 nm. Calculate the molar absorptivity of X .
"A = \u03b5bc"
Concentration of A = "\\dfrac{3.75 mg}{100 mL} \u00d7 \\dfrac{1000mL\/L}{335000mg\/mol} ="
"1.119 \u00d710^{-4} M"
"T = 0.396"
"A = - log T = - log 0.396 = 0.402"
"0.402 = \u03b5 \u00d7 1.50 cm \u00d7 1.119 \u00d710^{-4 }\\ mol\/L"
"\u03b5 = 2.4 \u00d7 10^{3} \\ L\/ mol. cm"
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