3A. Determine the Molarity of the Sodium ionin a solution of Na2SO4.10H20 if you add3.041 grams to a final volume of 150 ml.Add 3.041 g of Na2SO4.10H20 in a beakerpour upto 150ml of water to achieve 0.063Molar.CalculationNa2SO4.10H20 MW = 322.03g/molGrams = 3.041 gVf = 150ml = 0.15LM= mols of solute/ L of solution(3.041g/1) (1mol/ 332.2g) = 0.00915molNa2SO4.10H200.00915mol N22SO4.10H20/ .15l= 0.061 M2 Na = 2*0.061mol/L = 0.126 mol/LAnswer the question belowMake a 1:100 solution of Na2SO4.10H20 in100 ml of water from your previous solution.What is the final concentration of the Sodiumlon?

Given, Molarity of Na2SO4.10H2SO4 is 0.061M. 1:100 solution of Na2SO4.10H2O in 100 ml of water…

Answered: Make a 1:100 solution of Na2SO4.10H20…

Make a 1:100 solution of Na2SO4.10H20 in 100 ml of water from your previous solution. What is the final concentration of the Sodium lon?

Fundamentals Of Analytical Chemistry

9th Edition

ISBN:9781285640686

Author:Skoog

Publisher:Skoog

Chapter7: Statistical Data Treatment And Evaluation

Section: Chapter Questions

Problem 7.16QAP

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Question

3A. Determine the Molarity of the Sodium ion
in a solution of Na2SO4.10H20 if you add
3.041 grams to a final volume of 150 ml.
Add 3.041 g of Na2SO4.10H20 in a beaker
pour upto 150ml of water to achieve 0.063
Molar.
Calculation
Na2SO4.10H20 MW = 322.03g/mol
Grams = 3.041 g
Vf = 150ml = 0.15L
M= mols of solute/ L of solution
(3.041g/1) (1mol/ 332.2g) = 0.00915mol
Na2SO4.10H20
0.00915mol N22SO4.10H20/ .15l= 0.061 M
2 Na = 2*0.061mol/L = 0.126 mol/L
Answer the question below
Make a 1:100 solution of Na2SO4.10H20 in
100 ml of water from your previous solution.
What is the final concentration of the Sodium
lon?

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