moles . 0oiM Molanit 3. 3. A stock standard solution of 1.000x103 M Fe2* is prep

Question

moles . 0oiM Molanit 3. 3. A stock standard solution of 1.000x103 M Fe2* is prepared from ferrous ammonium sulfate, (NH4)2Fe(SO4)2. Standards A and B are prepared by adding with pipets 2,00 and 1.00 mL, respectively of the stock solution to 100-mL volumetric flasks and diluting to volume. Standards C, D, and E are prepared by adding 20.00, 10.00, and 5.00 mL of standard A to 100- mL volumetric flasks and diluting to volume. What are the concentrations of the five (5) working standard solutions? A 2.o -3 ,0002 D 1o.om A 0m

Explanation / Answer

1 a) Here your calculations are correct. 1.235 M is the answer

b) 0.895 m is the correct answer

2. Here, I couldn't follow what you, ve done. So, I have done my own calculation as below

0.9674+/-0.0009 g Na2CO3 x 1 mol / 105.9884+/-0.0007g =?

Error percentage= square root [(0.0009 / 0.9674 x100)2 + (0.0007 / 105.9884 x100)2] = 0.093034 %

Now,

(0.9674 /105.9884) = 0.009127+/-0.093034 mol Na2CO3

(0.009127 x 0.093034) = +/- 0.0008

0.009127+/-0.0008 mol Na2CO3 x 2mol HCl / 1mol Na2CO3 = 0.018254+/-0.0016 mol HCl

27.35 +/-0.04 mL HCl x 1L/1000 mL = 0.02735+/-0.00004 L HCl

0.018254+/-0.0016 mol HCl / 0.02735+/-0.00004L HCl =?

Error percentage = square root [(0.0016)2 + (0.00004/0.02735 x100)2] = 0.146%

0.018254 / 0.02735 = 0.667

Therefore, error percentage = (0.667 x 0.146) / 100= +/-0.001

The molarity of HCl = 0.667+/-0.001

3) Here, youve applied the equation CINITIAL x VINITIAL = CFINAL x VFINAL and all the claculations are correct

4)

a) Here the Buoyancy equation is correctly applied and 4.2391g is correct answer

b) The percentage is correctly calculated and 0.059 % is correct answer

Use comment box if further assistance required

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