NaOH solution would be in excess and thus prepare 1 M of HNO3 solution in burette, which will be used in back-titration. 4. Determine the end point of the back-titration when NaOH solution changes its color into pink. Record the results of at least three titrations. (Make a rough titration first).
Part C: Density of Sodium Chloride (NaCl) Solution, a sample of NaCl was obtained and measured using a 100mL beaker and a 10mL pipet to determine the concentration of the solution. In order to obtain the appropriate result, a calibration graph and density measurement was used to determine the concentration of the sodium chloride solution. In conclusion, based on the water temperature of 21.8°C in part A’s graduated cylinder experiment obtained, it was determined that the average density was .0973g/mL with a percentage error of 2.5%. When graphed the measurement was equal to Y=0.988x. Part B: The graduated pipet’s average density at 22.3 °C was determined to be 0.9785g/mL with a percentage error of 1.89% shows the graduated pipet to be more accurate and precise.
The following data were obtained when a sample of barium chloride hydrate was analyzed as described in the Procedure section. Calculate (a) the mass of the hydrate, (b) the mass of water lost during heating, and (c) the percent water in the hydrate. Mass of empty test tube 18.42 g Mass of test tube and hydrate (before heating) 20.75 g Mass of test tube and anhydrous salt (after heating) 20.41 g. Mass of the Hydrate is 2.33g. Loss (H2O) is 0.34g. Percent H2O in Hydrate is equal 0.34/2.33=14.6% 3.
Conclusion 10 Grams of Potassium chlorate when decomposed produces 3.915576 grams oxygen gas and 6.083363 grams potassium chloride Atomic Weight of Magnesium Introduction In this lab we will determine the atomic weight of magnesium by measuring the amount of hydrogen gas evolved when hydrochloric acid reacts with magnesium. The reaction is as follows: Mg + 2HCl -> H2 + Mg2+ (aq) + 2Cl- (aq) There is a one to one relationship between the number of moles of hydrogen gas evolved and the
3 x (C H5 N) = C3H15N3 Hydrated compounds Solving process: 1st- the difference between the initial mass and that of the dry sample is the mass of water that was driven off. Mass of hydrate minus mass of dry sample equals the mass of water 10.407 – 9.520 = 0.887 g 2nd- The mass of dry BaI2 and the mass of water are converted to MOLES. 9.520 g BaI2 x 1 mol BaI2 ∕ 391 g BaI2 = 0.0243 mol BaI2 anhydrate 0.887 g H2O x 1 mol H2O / 18.0 g H2O = o.o493 mol H2O 3rd: Dividing both results by the amt of 0.0243 mol, we get a ratio of 1 to 2.03, or 1 to 2, since the formula must have full numerical integers of water molecules, in other words no fractions of a water molecule. Thus, for every 1 mole of BaI2, there are two moles of water. The formula for the hydrate is written as BaI2 • 2H2O And it is named barium iodide dihydrate.
In the first part, five 100 mL flasks of 5 mL ligand solution, 5 mL 2 M sodium acetate, 4 mL 3 M NH2OH, and 1-5 mL Fe2+ solution are diluted with water. The absorption spectrum for varying concentrations of Fe2+ are measured using a spectrophotometer and the data is graphed in Excel. The slope of the line is ε in the Beer-Lambart equation A = εcl. In the second part of the experiment, eleven flasks containing diluted stock solutions of Fe2+ and ligand are mixed with 5 mL 2 M sodium acetate and 4 mL 3 M NH2OH and diluted with water. The absorption spectrum is measured using a spectrophotometer and the data is graphed in Excel.
From your three trials, calculate the average volume of Na2S2O3 needed for the titration of 25.00mL of diluted bleach. 3. Use the average volume and the molarity of Na2S2O3 to determine the molarity of the diluted bleach. (Find moles of Na2S2O3, convert to moles of NaClO, and divide by volume of dilute bleach that was titrated in each trial to get M). 4.
Calculate the molarity of the original vinegar solution and its concentration in gdm-3, given that it reacts with NaOH in a 1:1 ratio. 7. 2.5 g of a sample of ethanedioic acid, H2C2O4.nH2O, was dissolved in water and the solution made up to 250 cm3. This solution was placed in a burette and 15.8 cm3 were required to neutralise 25 cm3 of 0.1 moldm-3 NaOH. Given that ethanedioic acid reacts with NaOH
- Test tube - Distilled water - Volumetric flask. - LP4 tubes First 25ml of 0.1M Cupric Sulphate was made where the amount required was found in the following way... n=CV n-moles C-concentration V- Volume n= 0.1 x 25 x 10-3 =2.5 x 10-3 moles Then n = w/m n-moles w-mass required m- molecular weight 2.5 x 10-3 = w/249.71 molecular weight of CUSO4.5H2O=249.71 W = 0.624g Then as usual the solution was made by using the measured Cupric sulphate with the addition of water up to the point of 25ml in the volumetric flask. Then from the 0.1M stock solution of Cupric Sulphate the series of dilution was done as in the following table. The below mentioned volume of cupric sulphate was mixed with the corresponding volume of water as the final volume would be 2ml. The volume of cupric sulphate required for each concentration was calculated in the following way.