Trial 1 Trial3 Trial 2 31.5m. 2.0oml 29.Smi final buret reading 30.5m initial bu

Question

Trial 1 Trial3 Trial 2 31.5m. 2.0oml 29.Smi final buret reading 30.5m initial buret reading 29.5mt Volume oh NAOH used mL of standard acid used molarity of HCI molarity of NaOH Iomt 10m .302M Av. molality of NaoH B. Determination of Molar Mass: Sample No: 13 Trial 1 Trial 2 Trial 3 mass paper+ sample mass paper /25g lllg 125q Sooml 1.10ml 3 mL mass sample final buret reading initial buret reading Volume of NaOH used mL of standard acid added ml mol of HCI added mol of NaOH used mol of HCI reacted Molar mass. of oxide Av. Molar mass. Suggested formula for oxide (Show calculations on separate pages).

Explanation / Answer

Molarity of naOH can be calculated as follows: M1V1 = M2V2 where m1 and V1 is the concentration and volume of NaOH . M2 and V2 is the concentration and volume of HCl.

M1 = M2V2/V1

For trial 1 : M1 =0.1024 M

for trial 2 : M1 = 0.1024 M

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