please help find the other calculations with work so I can understand it! thank
ID: 713553 • Letter: P
Question
please help find the other calculations with work so I can understand it! thank you
Chem 1A Acid-base Titration B. Determination of the molar mass of an unknown solid UNKNOWN #: Trial 2 2.a32 3.185 225 3 Trial 1 1. Mass of weight boat (g) 2.932 2. Mass of weight boat + unknown sample (g) 2.15 252. Mass of weight boat + unknown sample (g) 2.1 3. Mass of unknown sample (g) 4. Average Molarity of NaOH (mol/L) S. Burette reading, initial (mL) 6. Burette reading, final (ml) 7. Volume of NaOH added (mL) 4162 8. Moles of NaOH added (mols) 9. Moles of unknown solid (mols) 10. Molar Mass of unknown (g/mol) 11. Average Molar Mass of Unknown (g/mol) 25.5 62 244 t17.63 Calculations:Explanation / Answer
Solution:- (8) We know that, molarity = moles/L
Trial 1, 0.338 = X/0.04162
X = 0.338 * 0.04162 = 0.01407 moles of NaOH
For trial 2) X = 0.338 * 0.04177 = 0.01412 moles
(9) Moles of unknonwn solid would be equal to the moles of NaOH used if the reaction is taking place in 1:1 mol ratio.
So, for traial 1) moles of unknown solid = 0.01407
for trial 2) moles of unknown solid = 0.01412
(10) Molar mass of unknonw solid = grams of unknown/moles of unknown
for trial 1) molar mass of unknown solid = 0.251g/0.01407 mol = 17.84 g/mol
for trial 2) molar mass of unknown solid = 0.253g/0.01412 mol = 17.92 g/mol
(11) Average molar mass of the unknown solid = (17.84g/mol + 17.92 g/mol)/2 = 17.88 g/mol
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