Good afternoon Chegg: In referencing Laboratory Manual Experiment 19. #5(b) A 10

Question

Good afternoon Chegg: In referencing Laboratory Manual Experiment 19. #5(b) A 10.0 mL sample of a saltwater solution containing sodium chloride, NaCl, was evaporated to dryness and gave the following data: mass of evaporating dish 51.925g, mass of evaporating dish 41.697g, and mass of evaporating dish and NaCl is 42.208g. Please show the calculation for the molar concentration of NaCl (58.44g/mol) in the saltwater sample. Please kindly show equations in written form. Thanks you for your help. Sincerely, How is 0.874 M the answer? Did you get the same answer?

Explanation / Answer

Sample of NaCl volume = 10 mL = 10 x 10-3 L

Mass of NaCl = 42.208 - 41.697 = 0.511 g

No. of moles of NaCl = 0.511g / 58.44 g mol-1 = 0.00874 mol

Molarity = No. of moles of NaCl / Volume of solution in L

= 0.00874 mol / 10 x 10-3 L

= 0.874 M

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