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What is the volume required to prepare 1.00 L of 100 x 10-3 M of Cl- out of 250 ppm stock solution of Mg2+ as MgCl2.

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Urine sample has a chloride concentration of 150 mN (meq/L). If all of the chloride in urine is assumed to be as NaCl; what is the concentration of NaCl in g/L.

please help me



Stock solution contains 250 mg Mg per 1000ml = 250ppm
Molecular mass of Mg Cl2 = 95.211
250 mg of Mg is contained in 95.211 / 24.3 times 250 mg of Mg Cl
= 979.5 mg Mg Cl2
which contains (35.45*2 / 95.211) times 979.5 mg of Cl
=729.3 mg of chlorine in 1000ml = 729.3 ppm Cl = (729.3/1000) / 35.45 molar
=0.0206 molar in Cl-

Dilution...
Assuming 1.00 x 10^-3 M is intended, not 100 x 10^-3..
(sign for indices is on the 6 key)

0.0206 M => 0.001 M

0.0206 / 0.001 = 20.6 ml of stock solution required, diluted to 1L

Stock solution contains 250 mg Mg per 1000ml = 250ppm
Molecular mass of Mg Cl2 = 95.211
250 mg of Mg is contained in 95.211 / 24.3 times 250 mg of Mg Cl
= 979.5 mg Mg Cl2
which contains (35.45*2 / 95.211) times 979.5 mg of Cl
=729.3 mg of chlorine in 1000ml = 729.3 ppm Cl = (729.3/1000) / 35.45 molar
=0.0206 molar in Cl-

Dilution...
Assuming 1.00 x 10^-3 M is intended, not 100 x 10^-3..
(sign for indices is on the 6 key)

0.0206 M => 0.001 M

0.0206 / 0.001 = 20.6 ml of stock solution required, diluted to 1L[/QUOTE]

Stock solution contains 250 mg Mg per 1000ml = 250ppm
Molecular mass of Mg Cl2 = 95.211
250 mg of Mg is contained in 95.211 / 24.3 times 250 mg of Mg Cl
= 979.5 mg Mg Cl2
which contains (35.45*2 / 95.211) times 979.5 mg of Cl
=729.3 mg of chlorine in 1000ml = 729.3 ppm Cl = (729.3/1000) / 35.45 molar
=0.0206 molar in Cl-

Dilution.....
Assuming 1.00 x 10^-3 M is intended, not 100 x 10^-3..
(sign for indices is on the 6 key)

0.0206 M => 0.001 M

0.0206 / 0.001 = 20.6 ml of stock solution required, diluted to 1L[/QUOTE]

This answer for What is the volume required to prepare 1.00 L of 100 x 10-3 M of Cl- out of 250 ppm stock solution of Mg2+ as MgCl2.

Ok, that's better.

Only the last part.

0.0206 mol Cl- is contained in 1000 mL

You need 0.001 mol of Cl-, diluted to 1 L.

0.0206 mol => 1000 mL
1 mol => 1000/0.0206 mL = 48600 mL
0.001 mol => 48600 * 0.001 = 48.6 mL

So, you need 48.6 mL of solution, diluted to 1 L.

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