REPORT FOR EXPERIMENT 22 (continued) NAME QUESTIONS AND PROBLEMS 1. If you had a
ID: 1047014 • Letter: R
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REPORT FOR EXPERIMENT 22 (continued) NAME QUESTIONS AND PROBLEMS 1. If you had added 50 mL of water to a sample of KHP instead of 30 ml, wouse? Explain tion of that sample then have required more, less, or the same amount of 2, A student weighed out 1.106 g of KHP How many moles was that? mol 3. A titration required 18.38 mL of 0.1574 M NaOH solution. How many moles of NaO were in this volume? mol 4. A student weighed a sample of KHP and found it weighed 1.276 g. Titration of this KHP required 19.84 mL of base (NaOH). Caleulate the molarity of the base. Forgetful Freddy weighed his KHP sample, but forgot to bring his report sheet along, so he recorded the mass of KHP on a paper towel. During his titration, which required 18.46 mL of base, he spilled some base on his hands. He remembered to wash his hands, but forgot about the data on the towel, and used it to dry his hands. When he went to calculate the molarity of his base, Freddy discovered that he didn't have the mass of his KHP. His kindhearted instructor told Freddy that his base was 0.2987 M. Calculate the mass of Freddy's KHP sample. 5. 6. What mass of solid NaOH would be needed to make 645 mL of Freddy's NaOH solution?Explanation / Answer
1) Addition of 50mL of water to a sample of KHP instead of 30mL , the titration of the sample will require the same amount of base. Because excess water will increase the volume of the sample number of moles of the sample remains same therefore it requires same amount of base.
2) Moles = weight / molar mass
Weight of sample = 1.106 g
Molar mass of KHP sample = 204.22 g/mol
Moles = 1.106 g / 204.22 g/mol
= 0.005415 mol
3) moles = concentration x volume ( volume =18.38 mL = 0.01838 L )
= 0.1574 x 0.01838
moles = 0.002893 mol
4)
Weight of KHP = 1.276 g
Molar mass of KHP sample = 204.22 g/mol
Moles = 1.276 g / 204.22 g/mol
= 0.006248 mol
Volume = 19.84 mL = 0.01984 L
Molarity = moles / volume
= 0.006248 mol / 0.01984 L
= 0.315 M
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