GRAND CANYON UNIVERSITY Data and Analysis Data and Analysis should be completed

Question

GRAND CANYON UNIVERSITY Data and Analysis Data and Analysis should be completed in the notebook. Include a section label, as well as labels for each Part and show one calculation for each type described. Part A 1. Create the following analysis table in the notebook labeled Part A:Molarity of NaC! Solutions. Use the questions below to calculate the values. Be sure to show a sample calculation for each type described. Part A: Molarity of NaCI Solutions Beaker # | Moles of NaCl solute | Molarity of Solution (mol/L) 1. Using the molar mass of NaCI (58.44 g/mol), and the grams of NaC1, calculate the moles of NaCl in each beaker. 2. Convert the volume of each solution from mL to L using a conversion factor. 3. Using the moles and the final volume of the solution (in L). determine the molarity of the solution in each beaker. ol Molarity of solution Moles of NaCI present in solution (moles) Total volume of solution (L) 4. Create a graph of density versus molarity for each of the solutions in the beakers and for the distilled water. The desity (g/mL) should go on the y-axis and the molarity (M) should go on the x-axis. Water will have a molarity of 0 M. Create a title for the graph and be sure to label each axis and include units. 5. On the graph, create a best fit line and display the equation of this line. A best-fit line is NOT connecting all the data points. Use Add Trendline and make sure that the linear option is selected. Select the Display equation on chart option. Do not select any other options including R? or Set itercept. 6. Print a copy of this graph and tape it into the notebook. © 2018, Grand Canyon University. All Rights Reserved.

Explanation / Answer

Worksheet 2

1.

Concentration = moles / volume of solution in litres = 5.8 mol / 0.25 L = 23.2 mol/L or 23.2 m

2.

Mass of Ca(NO3)2 = 45.0 g

molar mass = 164.09g/mol

moles of Ca(NO3)2 = 45.0g / 164.09g/mol = 0.27424 moles

1.3 M solution means 1.3 mol solute in 1 L solution

for 0.27424 moles solution needed = 1L/1.3 mol x 0.27424 mol = 0.21095 L = 210.95 mL

3.

molar mass of KNO3 = 101.10 g/mol

moles of KNO3 = molarity x volume = 0.355 M x 6.5L = 2.3075 moles

moles of KNO3 is also = mass / molar mass = m / 101.10 g/mol

equating both

2.3075 moles =  m / 101.10 g/mol

m = 233.29 g

4.

Na3PO4 will dissociate as:

Na3PO4 ------>3Na+(aq) + PO43-(aq)

17.5 M will give 3 x 17.5 = 52.5 M 17.5 M

Ans = Na+ = 52.5 M , PO43- = 17.5 M

i have answered top four only as not mentioned which one to solve

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