The resulting solution required 1221 ml of 01211 m naoh to


In one experiment, a student added a total of 15.00 mL of 0.5581 M HCl to 1.4766 g to of an unknown which contained CaCO3 (molar mass = 100.09 g/mol). The resulting solution required 12.21 mL of 0.1211 M NaOH to back titrate the excess HCl. What was the percent CaCO3 in the unknown? Report your answer to two (2) decimal places

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Chemistry: The resulting solution required 1221 ml of 01211 m naoh to
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