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Suppose that you experimentally determine the molar extinction coefficient (ε) of a complex ion sample to be 8171 L/mol-cm at 510 nm. The literature value for the ε of the complex ion sample is reported as 8400 L/mol-cm. What is the percent purity of your experimental sample

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The percent purity of your experimental sample is 97.3 %

The percentage purity % = impure value/pure value × 100 %.

Actual value of molar extinction coefficient = impure value = 8171 L/mol-cm, pure value = literature value = 8400 L/mol-cm.

Substituting the values of the variables into the equation, we have

% purity = 8171 L/mol-cm/8400 L/mol-cm ×  100 %

% purity = 0.9727 × 100 %

% purity =  97.27 %

% purity ≅  97.3 %

The percent purity of your experimental sample is 97.3 %

Learn more about percentage purity here:

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