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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
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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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
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.
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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