G.23 from HW  [ENDORSED]

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MeghetyManoyan1A
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Joined: Fri Apr 06, 2018 11:03 am

G.23 from HW

Postby MeghetyManoyan1A » Sun Apr 08, 2018 4:05 pm

For problem G.23,

"In medicine it is sometimes necessary to prepare solutions with a specific concentration of a given ion. A lab technician has made up a 100.0-mL solution containing 0.50 g of NaCl and 0.30 g of KCl, as well as glucose and other sugars. What is the concentration of chloride ions in the solution?,"

I am confused on how to find the concentration of only the chloride ions. Do we find separate concentrations for NaCl and KCl?

Kalsuda Lapborisuth 1B
Posts: 27
Joined: Fri Apr 06, 2018 11:03 am

Re: G.23 from HW  [ENDORSED]

Postby Kalsuda Lapborisuth 1B » Sun Apr 08, 2018 5:08 pm

Because when KCl and NaCl dissolves in water, they become K+and Cl- and Na+ and Cl- respectively, you can approach this problem by simply adding the moles of KCl and NaCl together to find the total molar concentration of Cl- in 100.0mL solution.

AshleyLamba1H
Posts: 30
Joined: Fri Apr 06, 2018 11:03 am

Re: G.23 from HW

Postby AshleyLamba1H » Sun Apr 08, 2018 6:14 pm

The units for molar concentration is mol/L. Therefore, you must first convert the amount of grams of Cl- to moles, and then divide that by .100 L (100 mL).

jadam_1E
Posts: 30
Joined: Fri Apr 06, 2018 11:01 am

Re: G.23 from HW

Postby jadam_1E » Sun Apr 08, 2018 11:01 pm

For this problem, I approached it by thinking that for NaCl, the proportion of Cl ions to NaCl formula units is 1:1, and the proportion of Cl atoms to KCl formula units is also 1:1. Therefore, the concentration of Cl ions in the solution will be the same as the concentration calculated by converting .50g NaCl and .30g KCl to moles, adding together these separate number of moles, and dividing by the given volume of the solution which is 100mL.

Ani Boyadjian 1A
Posts: 30
Joined: Fri Apr 06, 2018 11:02 am

Re: G.23 from HW

Postby Ani Boyadjian 1A » Sun Apr 15, 2018 2:02 pm

I first converted both NaCl and KCl to moles, then added them together and finally multiplied by (1/0.1L)


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