By using the data page complete the results page Bata Page Name Part A-Standardi

Question

By using the data page complete the results page Bata Page Name Part A-Standardization of NaoH solution Molarity of standardized Hc (provided by your instructor) Dsla1b Trial Iria12 Initial Reading, HCI burette se ee mL. 20 mL Final Reading HCl burette 2a Eo mu mL Initial Reading, NaOH burette or mL. mL Final Reading, NaoH burette 3 Loo mL. eg,so mL Part B Determination of Molar Mass of an Unknown Acid Unknown number Mass of sample 1. o.20 6.24 grams Mass of sample 2 Q2 grams Iria 1 Trial 2 mL. mL. Initial Reading, NaOH burette mL. mL. Final Reading, NaOH burette

Explanation / Answer

Part A

Mnaoh = 0.0176×19.90/30.99 = 0.0113

Now

M1V1=MnaohVnaoh

VHCL= 0.0113×30.99/0.0176 =19.89

Trial 2

Mnaoh= 0.0176×20/29.80= 0.0118

VHcl = 0.0118× 29.80/0.0176 = 19.97

Total volume of NaoH

Trial 1

Mnaoh × Vnaoh = Mhcl×V

Volume of NaoH = 0.0176×19.90/0.0113 = 30.99

Trial 2

Volume of Naoh = 0.0176×20/0.0113 = 31.15

Trail 1 T2

Molarity of naoh 0.0113 0.0118

mean = 0.0172

Part B T1 T2

mass of sample 0.20 0.20

volume of naoh used 30.90 30.80

moles of naoh used in trial 1

0.0172×30.90/1000 = 0.00053 Moles

moles of naoh = Moles of Acid

thus

0.00053 Moles of Acid = 0.20 gm of acid

one mole of acid = 0.20/0.00053 = 377.35

Trial 2

moles of naoh used = 0.0172×30.80/1000= 0.00052

one mole of acid = .20/0.00052= 384.61

average molar mass = 380.5

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