What is the Molarity of 1.00 ppm of Fe2+ prepared from ammonium sulphate (FeSO4.(NH4)2SO4.6H2O).

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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.

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.

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Tris(hydroxymethyl)aminomethane (THAM) is a weak base frequently used to prepare buffers in biological work since the corresponding pKa is 8.08, which is near the pH of the physiological buffers. What is the weight of THAM required to prepare a 1 L buffer with pH 7.40 and 100 mL of 0.50 M HCl.


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What is the pH of the buffer produced by mixing 5.0 mL of 0.1 M NH3 with 10 mL of 0.020 M HCl. Determine the pH after 1:1 dilution or addition of 1mL of 1.0 mM NaOH or HCl.

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Quote:

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Originally Posted by zezo254

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

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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]

Quote:

---

Originally Posted by zezo254

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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