Question

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Goud to CAT CATEGORY A SS BACK TYTRATION N 13-20_43.81 SOL 3TTE 10 On approximatymarket carbs y heder 250 ml A 300 M othek and w e re complete and the ple company dissolved in my heart beklager with a congredo speed up the pas the solution does not find as do 200 of Anthelio danga p hen their is you in and in base en perso al baliquot of 2 Add approximately 20 ml. delo w a 3A 100 500 MHChe m ical and temple is complewy dissolved. It may be m ary large chunks up with a singud toond up the pres If the sidesp e n al ml linaw If the solution apps to hear , add an additional 50 m. lique HCL T he added to the end Report the promotie EU) 1264CO2-Weos OH Materih ERZH4y03->2010, +2N4 SL Volpip e : HC OOS->120 +by+N4 05 27.12 SOL volume of NCH THOSE EGGS WAR Gver rimerkit HYK NEDED mandardized i p Write a bal cution for the action betwem your carbone and HC l ar and # S> GRATIS GENCH POLARITT I BOTTLE Hal = 0.5M IS O. 1044 M N901 Marity of Norts Carbonate LOZ RACCO 1 Mes of Weibout 2 Mass of Wig Mas of Wegboml • Mass of Carbon 5 sme of Moles of HCI 3 - GRAMS

Answer 1

Part A

1. Equation#2 and Equation#3 as written by you are the answers to this question

2. (b) Molarity of HCl solution: Already given, i.e. 0.5 M

(c) Molarity of NaOH solution: As mentioned on your bottle (which was previously standardized), i.e. 0.1044 M

6. Moles of HCl = 0.005 moles

Explanation:

Molarity of HCl solution = 0.5 M = 0.5 mol/L = 0.5 mol in 1000ml

Volume of HCl added (as in experiment) = 10 ml

Therefore moles of HCl = 0.5/1000*10 = 0.5/100 = 0.005 moles

10. Moles of NaOH dispensed = 0.0038 moles

Explanation:

Molarity of NaOH solution = 0.5 M = 0.1044 mol/L = 0.1044 mol in 1000ml

Volume of NaOH dispensed (as in trial 1) = 36.3 ml

Therefore moles of NaOH dispensed = 0.1044/1000*36.3  = 0.0038 moles

11. Moles of carbonate = 0.0006 moles

Explanation:

For every 2 mole of HCl consumed, 1. mole of carbonate reacts as per the equation#2.

But moles of HCl consumed = moles of HCl added - moles of HCl unreacted

Moles of HCl unreacted = moles of NaOH dispensed = 0.0038 moles

Thus, Moles of carbonate = moles of HCl consumed/2 = (0.005 - 0.0038)/2 = 0.0006 moles

12. Formula mass of carbonate =

Explanation:

Formula mass of carbonate = Mass of carbonate added to the flask/ Moles of carbonate

= 0.293/0.0006

= 488.33 g/mol

13. Average formula mass of carbonate = Repeat the above calculations of trial 2 and average the value of formula mass of carbonate that you will obtain from trial 2

14. The data is to be provided by you

15. Without the data 14 available, 15 cannot be solved

Part B:

All calculations are same as part A. you may follow the same methodology and get correct answers.

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