Santa Monica College Chemistry 1 Another student decides to use the technique em
ID: 569413 • Letter: S
Question
Santa Monica College Chemistry 1 Another student decides to use the technique employed in Part C of this lab to determine the density of gold. She obtains the following data: 3. (A) Mass of Empty Vial (B) Mass of Vial+ Gold Sample (C) Mass of Vial + Gold+ Water (D) Mass of Vial+ Water (E) Temperature of Water F) Density of Water at above Temp.0.9970 g/ 10.495 g 70.773 g 76.288 g 19.119 g 25.0 C Using her data, perform the calculations below. Show your work clearly, and pay attention to significant figures and units. a) What is the mass of the gold sample? What is the mass of water in the vial with the gold (see measurement C)? What is volume of water in this vial (hint: use the density of water)? b) o) What is the mass of water in the vial filled with water only (see measurement D)? What is the volume of water in this filled vial (hint: use the density of water)? d) The volume of the gold is equal to the difference in the two water volumes. What is the volume of the gold? e) Determine the density of the gold, in g/cm Page 2 of 2 The Densities of Solutions and SolidsExplanation / Answer
Ans. #a. Mass of gold sample =
(mass of empty vial + Gold sample) – (mass of empty vial)
= 70.773 g – 10.495 g
= 60.278 g
#b. Mass of water in the vial with gold sample =
(mass of empty vial + Gold + water) – (mass of empty vial + Gold sample)
= 76.286 g – 70.773 g
= 5.513 g
# Volume of water = Mass / Density = 5.513 g / (0.9970 g/ mL) = 5.530 mL
#c. Mass of water in the vial filled with water>
(mass of empty vial + water) – (mass of empty vial)
= 19.119 g – 10.495 g
= 8.624 g
# Volume of water = Mass / Density = 8.624 g / (0.9970 g/ mL) = 8.650 mL
#d. Volume of gold = Calculated water volume is #c – Calculated water volume in #d
= 8.650 mL - 5.530 mL
= 3.120 mL
#e. So far, we have-
Mass of gold sample = 60.278 g
Volume of gold = 3.120 mL
Now,
Density of gold = Mass / Volume
= 60.278 g / 3.120 mL
= 19.320 g/ mL
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