Part B A student placed 18.5 g of glucose (C6 H12 O6) in a volumetric flask, added enough water to dissolve the glucose by swirling, then carefully added additional water until the 100. mL mark on the neck of the flask was reached. The flask was then shaken until the solution was uniform. A 20.0 mL sample of this glucose solution was diluted to 0.500 L. How many grams of glucose are in 100. mL of the final solution? Express your answer to three significant figures and include the appropriate units. • View Available Hint(s) "i HẢ ? Value Units
Part B A student placed 18.5 g of glucose (C6 H12 O6) in a volumetric flask, added enough water to dissolve the glucose by swirling, then carefully added additional water until the 100. mL mark on the neck of the flask was reached. The flask was then shaken until the solution was uniform. A 20.0 mL sample of this glucose solution was diluted to 0.500 L. How many grams of glucose are in 100. mL of the final solution? Express your answer to three significant figures and include the appropriate units. • View Available Hint(s) "i HẢ ? Value Units
Chemistry: An Atoms First Approach
2nd Edition
ISBN:9781305079243
Author:Steven S. Zumdahl, Susan A. Zumdahl
Publisher:Steven S. Zumdahl, Susan A. Zumdahl
Chapter6: Types Of Chemical Reactions And Solution Stoichiometry
Section: Chapter Questions
Problem 123CP: The units of parts per million (ppm) and parts per billion (ppb) are commonly used by environmental...
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Please answer question 2 Part A and B
![Part B
A student placed 18.5 g of glucose (C6 H12O6) in a volumetric flask, added enough water to dissolve the glucose by
swirling, then carefully added additional water until the 100. mL mark on the neck of the flask was reached. The flask
was then shaken until the solution was uniform. A 20.0 mL sample of this glucose solution was diluted to 0.500 L.
How many grams of glucose are in 100. mL of the final solution?
Express your answer to three significant figures and include the appropriate units.
• View Available Hint(s)
HÀ
?
Value
Units](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc6ad8812-3da5-4be0-a483-33eefd12dc82%2F9860dad9-81db-41d4-8056-168736e9480d%2F545yvrw_processed.png&w=3840&q=75)
Transcribed Image Text:Part B
A student placed 18.5 g of glucose (C6 H12O6) in a volumetric flask, added enough water to dissolve the glucose by
swirling, then carefully added additional water until the 100. mL mark on the neck of the flask was reached. The flask
was then shaken until the solution was uniform. A 20.0 mL sample of this glucose solution was diluted to 0.500 L.
How many grams of glucose are in 100. mL of the final solution?
Express your answer to three significant figures and include the appropriate units.
• View Available Hint(s)
HÀ
?
Value
Units
![Molarity (M) is defined as the number of moles of solute
divided by the solution volume expressed in liters:
Part A
moles of solute
molarity =
%3D
volume of solution (L)
The following five beakers, each containing a solution of sodium chloride (NaCl, also known as table salt), were
For example, 1 M HCl contains 1 mol of HCI
dissolved in 1 L of solution. When a concentrated
found on a lab shelf:
solution is diluted, the number of moles of solute stays
constant; only the volume of the solution is changed. A
dilution indicates an increase in solution volume and,
therefore, the concentration of the solution must
decrease. If you add more water to the HCl solution
considered above, so that now the volume is 2 L , the
Beaker
Contents
1
200. mL of 1.50 M NaCl solution
100. mL of 3.00 M NaCl solution
3
150. mL of solution containing 21.0 g of NaCl
number of moles remains the same but the volume is
4
100. mL of solution containing 21.0 g of NaCl
doubled. Hence the molarity of the solution is now 1 mol
in a 2 L solution, that is, (1/2) M or 0.5 M.
5
300. mL of solution containing 0.450 mol NaCi
The number of moles of solute before and after dilution
Arrange the solutions in order of decreasing concentration.
can be calculated by multiplying molarity times volume.
We can set up the following equations:
Rank from most concentrated to least concentrated. To rank items as equivalent, overlap them.
moles of solute
molarity x volume
• View Available Hint(s)
M; x V = Mt X Vị
where Mi is the initial molarity (of the concentrated
solution), Vị is the initial volume, Mf is the final molarity
(of the diluted solution), and Vị is the final volume.
Reset
Help
In the HCl solution example the initial molarity is 1 M,
the initial volume is 1 L, the final volume is 2 L, and the
molarity is 0.5 M. Thus the number of moles present in
| 4] 2][3] 51]
these solutions is
M; V = Mt V = 1 M × 1 L = 0.5 M × 2 L
1 mol
Most concentrated
Least concentrated
2.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc6ad8812-3da5-4be0-a483-33eefd12dc82%2F9860dad9-81db-41d4-8056-168736e9480d%2Ff6aqg1m_processed.png&w=3840&q=75)
Transcribed Image Text:Molarity (M) is defined as the number of moles of solute
divided by the solution volume expressed in liters:
Part A
moles of solute
molarity =
%3D
volume of solution (L)
The following five beakers, each containing a solution of sodium chloride (NaCl, also known as table salt), were
For example, 1 M HCl contains 1 mol of HCI
dissolved in 1 L of solution. When a concentrated
found on a lab shelf:
solution is diluted, the number of moles of solute stays
constant; only the volume of the solution is changed. A
dilution indicates an increase in solution volume and,
therefore, the concentration of the solution must
decrease. If you add more water to the HCl solution
considered above, so that now the volume is 2 L , the
Beaker
Contents
1
200. mL of 1.50 M NaCl solution
100. mL of 3.00 M NaCl solution
3
150. mL of solution containing 21.0 g of NaCl
number of moles remains the same but the volume is
4
100. mL of solution containing 21.0 g of NaCl
doubled. Hence the molarity of the solution is now 1 mol
in a 2 L solution, that is, (1/2) M or 0.5 M.
5
300. mL of solution containing 0.450 mol NaCi
The number of moles of solute before and after dilution
Arrange the solutions in order of decreasing concentration.
can be calculated by multiplying molarity times volume.
We can set up the following equations:
Rank from most concentrated to least concentrated. To rank items as equivalent, overlap them.
moles of solute
molarity x volume
• View Available Hint(s)
M; x V = Mt X Vị
where Mi is the initial molarity (of the concentrated
solution), Vị is the initial volume, Mf is the final molarity
(of the diluted solution), and Vị is the final volume.
Reset
Help
In the HCl solution example the initial molarity is 1 M,
the initial volume is 1 L, the final volume is 2 L, and the
molarity is 0.5 M. Thus the number of moles present in
| 4] 2][3] 51]
these solutions is
M; V = Mt V = 1 M × 1 L = 0.5 M × 2 L
1 mol
Most concentrated
Least concentrated
2.
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