I\'m having trouble figuring out how to do the calculations. Can someone please

Question

I'm having trouble figuring out how to do the calculations. Can someone please help me! 25.00 mL HCL was added and the molar concentration is .1M HCl

Experiment 17 Report Sheet Antacid Analysis A g the Antacia 1. Mas of Sask () 3. Mass (or tured mans) of antacid anple () 4 Tuital volame of HC added (nl) S. Molar conceotration of HCTT) B. Analyzing the Antacid Sample L. Molar concentration of NaOH(mo 3 Baret reading,Foal (nL) 4 Volame of NaOH (ml C. Calculations L Moles of HC) added, otal (mol) 2. Moles of NaOH added ( -3. Miles of base in antacid sample (mo) mol base in antacid (NoUg) mass of sntacid sple) mol base in antacid(v) 5. Aerags mass of astackd sample Show ealculation(s) for Trial(s) 1 on the next page.

Explanation / Answer

Outline to solve the problem:

Molarity (M) = no. of mol / volume (L)

i.e. M = no. of mmol * 1000 / (Volume (mL) * 1000)

i.e. M = mmol/mL

According to the given data, the no. of mmol of HCl = 2.5 mmmol

The volume of HCl = 25 mL

Now, the molarity (M) = 2.5 mmol / 25 mL = 0.1 mmol/mL = 0.1 M

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