A volume of 13.96 mL of 0.1060 M NaOH solution was used to titrate a 0.618...

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Chemistry

A volume of 13.96 mL of 0.1060 M NaOH solution was used totitrate a 0.618 g sample of unknown containingHC₇H₅O₃.

What is the molecular mass ofHC₇H₅O₃? (report answers to 4 or 5significant figures) 1.3812Ã—10²g/mol

What is the percent by mass ofHC₇H₅O₃ in the unknown?

In this problem what mass of sample in grams would be needed todeliver about 23.40 mL in the next trial?

In the second trial above, exactly 1.032 g was transferred into aflask to be titrated. If the initial buret reading is 0.10 mL,predict what the final buret reading be. ?

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3.9 Ratings (461 Votes)

STEP 1 CALCULATION OF NUMBER OF MOLES OF NAOH IN 1396 mL of 01060 M NaOH SOLUTION 1000 mL of 1 M NaOH contains 1 mol NaOH 1396 mL of 01060 M NaOH contains 1 x 1396 x 01060 1000 x 1 000148 mol NaOH STEP 2 CALCULATION OF NUMBER OF MOLES OF HC7H5O3 IN 0618 g sample of unknown containing HC7H5O3 HC7H5O3 is monoprotic monobasic acid Note that one H atom is wrtten separately Hence See Answer

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