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13-3. How would you prepare 50.0-mL portions of SS that are 0.00500 M, 0.00200 M and 0.00100 M in Na* from the soln in Ex. 2?

13-3. How would you prepare 50.0-mL portions of SS that are 0.00500 M, 0.00200 M and 0.00100 M in Na* from the soln in Ex. 2?

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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EX 2.

Given->

Volume of Na+ = 500 ml

Molarity of Na+= 0.0100M

Molar mass of Na2CO3 = 105.99 gm/mole

 

Millimole of Na+ = molarity × volume

Number of millimole = 0.0100 × 500

= 5 millimole

Na2CO3 ---> 2Na+ + CO32-

Millimole of Na2CO3 =  millimole of Na+/2

Millimole of Na2CO3 =5/2 = 2.5 millimole

Mole of Na2CO3 = 2.5 × 10-3mole (1 mole = 10^3 millimole)

Weight of Na2CO3 required = mole × molar mass

= 2.5 × 10-3 × 105.99

=0.26 gm

 

Hence, 0.26 gm Na2CO3 must dissolve in 500 ml of water.

13-3. How would you prepare 50.0-mL portions of SS that are 0.00500 M, 0.00200 M and 0.00100 M in Na* from the soln in Ex. 2?
Transcribed Image Text:13-3. How would you prepare 50.0-mL portions of SS that are 0.00500 M, 0.00200 M and 0.00100 M in Na* from the soln in Ex. 2?
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