Part B In Part A, you explored the relationship between molarity, the number of
ID: 1006483 • Letter: P
Question
Part B
In Part A, you explored the relationship between molarity, the number of moles, and solution volume. However, when you prepare a solution, you are unable to directly measure the number of moles of solute. Instead, you can mass the solute and convert between the number of moles and mass using the molar mass of a substance as a conversion factor. The molar mass of a substance can be calculated based on its molecular formula; otherwise, it can be calculated from the mass and number of moles as follows:
Molar Mass = Mass solute (in g)Moles solute
In order to prepare a certain volume of solution with a desired concentration, you would need to determine the required mass of solute. What mass of MgCl2 would you need to prepare a 100. mL solution at a concentration of 0.390 M ?
Express the mass in grams to three significant figures.......
Part C
You need to prepare a 1.90 M solution of potassium chloride (molar mass of potassium chloride = 74.55 g/mol ), but you only have a 10 mL graduated cylinder and a 25 mL beaker. Complete the following sentences regarding the concentration of the prepared solution.
Match the words in the left column to the appropriate blanks in the sentences on the right. Make certain each sentence is complete before submitting your answer.
1. In order to prepare 10.0 mL (0.0100 L) solution of 1.90 M potassium chloride, a mass of ________ would be required.
2. When the solute and a total of 10.0 mL of water are both added into the beaker and stirred until the solute dissolves, the solution volume will be ________ .
3. Based on that, the concentration of the prepared solution will be __________ .
OPTIONS :
0.019 g
1.42 g
exactly 10.0 mL
less than 1.90 M.
less than 10.0 mL
exactly 1.90 M
Part D
You are asked to prepare a 1.000 L solution of 4.5 M C6H12O6 (glucose; molar mass = 180.16 g/mol) in a lab by dissolving 811.0 g of glucose in water. Consider the following two scenarios in which you commit a user error while preparing this solution.
You decide to evaluate and compare the errors you made while preparing the solutions using the different methods. Calculate the actual concentrations of the intended 4.5 M glucose solutions prepared by each method based on their actual final volumes.
Express the concentrations in molarity to two decimal places. Make certain each field is complete before submitting your answer.
Flask g glucose
flask two = concentraion: _____M
flask three 811.0 g glucose
flask four= concentration;_______M
greater than 10.0 mL
greater than 1.90 M
Prepared in a volumetric flask Prepared in a beaker Assumed volume 1.000 L 1.000 L Volumetric error Added water 2.0 cm above the line, which corresponds to 6.3 mL (0.0063 L) additional solution volume 811.0 g takes up 526 mL of space rather than 500 mL of space Preparation details You add the glucose to a volumetric flask and then add water until it dissolves. The water bottle you are using has a worn tip, and you inadvertently add too much water such that the meniscus is above the line. The diameter of the neck of the volumetric flask is 2.29 cm. You prepare the solution by adding glucose to a large beaker and approximated the volume of the glucose to be 500 mL . Therefore, you add 500. mL of water to the 526mL in the beaker using a graduated cylinder.Explanation / Answer
PART B:
CONCEPT: Use the following equation to solve the problem
M = n / V
M = molarity = 0.390M
n = number of moles = Mass of solute / Molar mass = m / 95.2
V = volume of solution in liters = 100 / 1000 = 0.1L ; Volume = 100mL = 100mL /1000 = 0.1L
0.390 = (m / 95.2 X 0.1L )
m = 0.390 X 95.2 X 0.1L = 3.71g
PART D:
1. Use the above equation to find the answer
1.90 = m / 74.55 X 0.01; m = 1.42g
2. Exactly 10mL
3. Exactly 1.9M
PART D:
(i) Concentration in Volumetric flask
M = (811.0 / 180.16 x 1.0063) = 4.47M
(ii) Concentration in beaker
M = (811.0 / 180.16 x 1.026) = 4.388M