Discussion This virtual experiment involves the determination of the concentration of salt (sodium chloride, NaCl) in an aqueous solution. Concentration may be expressed many ways, and this experiment studies two systems: molarity (M) and weight/volume percent (%w/v). Molarity is defined as the number of moles of solute (in this case NaCl) in a liter of solution. It is calculated with the following equation: M = moles solute liters solution Percent concentration (w/v) is defined as the number of grams of solute (in this case NaCl) in 100. mL of solution. It is calculated with the following equation: %(w/v) = grams solute x 100 mL solution The experimental method involves taking a specific volume of an aqueous salt solution and slowly evaporating it to dryness. The weight which remains is that of the solute. From the weight of solute and the initial volume of solution, both molarity and percent (w/v) can be calculated. Experimental in this virtual experiment, three students (A, B and C) determined the salt content of three identical samples. All the data that would have been collected in an actual experiment are recorded on the attached report form. Two concentrations, molarity and %(w/v), of salt in 5.00 mL of solution were determined based on the weight the dry salt and the volume of the solution. Calculations The student data is entered in the table on the next page. Based on their data, you are required to enter all the calculated values in the spaces provided in the report form. Calculate the weight of the salt by subtracting the tare weight of the evaporating dish from its final dry weight Convert the weight of sodium chloride into moles. Convert the volume of solution into liters. Calculate the molarity of the solution from the moles and liters by substituting these values into the equation for M. Calculate the percent (w/v) of a solution made by dissolving 23.4 g sodium chloride in water to make a solution with a volume of 250. mLCalculate the number of grams of NaOH needed to make 1500. mL of a 0.200 M solution.0 mL of a potassium chloride solution is transferred to an evaporating dish and the water evaporated. The weight of KCl is 0.75 g.calculate the molarityCalculate the percent sodium chloride (w/v) from the weight of sodium chloride and the volume of solution by substituting these values into the equation for %(w/v). Complete the Following Exercises Calculate the number of grams of NaOH needed to make 1500. mL of a 0.200 M solution.

Discussion This virtual experiment involves the determination of the concentration of salt (sodium chloride, NaCl) in an aqueous solution. Concentration may be expressed many ways, and this experiment studies two systems: molarity (M) and weight/volume percent (%w/v). Molarity is defined as the number of moles of solute (in this case NaCl) in a liter of solution. It is calculated with the following equation: M = moles solute liters solution Percent concentration (w/v) is defined as the number of grams of solute (in this case NaCl) in 100. mL of solution. It is calculated with the following equation: %(w/v) = grams solute x 100 mL solution The experimental method involves taking a specific volume of an aqueous salt solution and slowly evaporating it to dryness. The weight which remains is that of the solute. From the weight of solute and the initial volume of solution, both molarity and percent (w/v) can be calculated. Experimental in this virtual experiment, three students (A, B and C) determined the salt content of three identical samples. All the data that would have been collected in an actual experiment are recorded on the attached report form. Two concentrations, molarity and %(w/v), of salt in 5.00 mL of solution were determined based on the weight the dry salt and the volume of the solution. Calculations The student data is entered in the table on the next page. Based on their data, you are required to enter all the calculated values in the spaces provided in the report form. Calculate the weight of the salt by subtracting the tare weight of the evaporating dish from its final dry weight Convert the weight of sodium chloride into moles. Convert the volume of solution into liters. Calculate the molarity of the solution from the moles and liters by substituting these values into the equation for M. Calculate the percent (w/v) of a solution made by dissolving 23.4 g sodium chloride in water to make a solution with a volume of 250. mLCalculate the number of grams of NaOH needed to make 1500. mL of a 0.200 M solution.0 mL of a potassium chloride solution is transferred to an evaporating dish and the water evaporated. The weight of KCl is 0.75 g.calculate the molarityCalculate the percent sodium chloride (w/v) from the weight of sodium chloride and the volume of solution by substituting these values into the equation for %(w/v). Complete the Following Exercises Calculate the number of grams of NaOH needed to make 1500. mL of a 0.200 M solution.

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...

See similar textbooks

Related questions

Concept explainers

Question

Discussion

This virtual experiment involves the determination of the concentration of salt (sodium chloride, NaCl) in an aqueous solution. Concentration may be expressed many ways, and this experiment studies two systems: molarity (M) and weight/volume percent (%w/v).

Molarity is defined as the number of moles of solute (in this case NaCl) in a liter of solution. It is calculated with the following equation:

M = moles solute
liters solution

Percent concentration (w/v) is defined as the number of grams of solute (in this case NaCl) in 100. mL of solution. It is calculated with the following equation:

%(w/v) = grams solute x 100

mL solution

The experimental method involves taking a specific volume of an aqueous salt solution and slowly evaporating it to dryness. The weight which remains is that of the solute. From the weight of solute and the initial volume of solution, both molarity and percent (w/v) can be calculated.

Experimental

in this virtual experiment, three students (A, B and C) determined the salt content of three identical samples. All the data that would have been collected in an actual experiment are recorded on the attached report form. Two concentrations, molarity and %(w/v), of salt in 5.00 mL of solution were determined based on the weight the dry salt and the volume of the solution.

Calculations

The student data is entered in the table on the next page. Based on their data, you are required to enter all the calculated values in the spaces provided in the report form.

Calculate the weight of the salt by subtracting the tare weight of the evaporating dish from its final dry weight
Convert the weight of sodium chloride into moles.
Convert the volume of solution into liters.
Calculate the molarity of the solution from the moles and liters by substituting these values into the equation for M.
1. 1. 1. 1. Calculate the percent (w/v) of a solution made by dissolving 23.4 g sodium chloride in water to make a solution with a volume of 250. mLCalculate the number of grams of NaOH needed to make 1500. mL of a 0.200 M solution.0 mL of a potassium chloride solution is transferred to an evaporating dish and the water evaporated. The weight of KCl is 0.75 g.calculate the molarityCalculate the percent sodium chloride (w/v) from the weight of sodium chloride and the volume of solution by substituting these values into the equation for %(w/v).
        Complete the Following Exercises
        
        Calculate the number of grams of NaOH needed to make 1500. mL of a 0.200 M solution.

Expert Solution

Trending now

This is a popular solution!

Step by step

Solved in 2 steps with 1 images

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.