The figure calibrates absorbance against chromate concentration ($CrO_4^{2-}$,mg/L). For sample absorbance .35,find Cr(VI) concentration in mg/L, to two decimal places. Atomic weights are Cr 52 and O 16 g/mol.
The calibration is $y=.8003x+.0055$,so chromate concentration is $x=(.35-.0055)/.8003=.4304636$ mg/L. Each mole of chromate contains one mole of chromium, and its formula mass is $52+4(16)=116$. Thus chromium concentration is $.4304636(52/116)=.1929664$ mg/L, rounded **0.19 mg/L**. The calibration’s chromate concentration must not be reported directly as chromium.
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