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# Chemistry: Ammonium Meta Tungstate

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AMT (ammonium meta tungstate) has a MW of 2956.2 g/mol. Calculate the amount of AMT (in grams) needed to be dissolved in water that would give a 0.2 M AMT aqueous solution.

Also on an unrelated problem, how would you dilute (with water) 12.1 Normality HCl and 14.8 Normality NH4H2O to give 1M HCL and 1M NH4H20 respectively?

https://brainmass.com/chemistry/acids-and-bases/chemistry-ammonium-meta-tungstate-405850

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Question 1:

AMT (ammonium meta tungstate) has a MW of 2956.2 g/mol. Calculate the amount of AMT (in grams) needed to be dissolved in water that would give a 0.2 M AMT aqueous solution.

The molar concentration of a solution (M) represents the number of moles dissolved in one liter of solution (usually water). So an 0.2M solution of AMT would contain 0.2 moles in 1 liter of water.

The molecular weight of a compound or element is calculated from it's chemical formula (sum of the atomic weights of the constituent atoms). The molecular weight of ammonium meta tungstate (NH4)6H2W12O40 is calculated by summing the individual atomic weights (2956.2 amu).

The molecular weight of a single atom is small* (measured in atomic mass units). It is more convenient to express mass and molecular weight in grams. In order to do this the concept of the mole was developed. A mole is the number of atoms (6.02x10^23: Avogadro's number) that combined equal its molecular (or atomic weight) in grams. For example carbon has an atomic weight weight of 12 amu. 1 mole (or 6.02x10^23 atoms) weighs 12 g. This is why the molecular weight of ammonium meta tungstate is expressed as 2956.2 g/mol

Getting back to ammonium meta tungstate an 0.2M solution would contain 0.2 mol/liter x 2956.2 g/mol or 591.24 g dissolved in 1 liter of water

* 1 amu = 1.66X10^-24 grams

Question 2:

.. how would you dilute (with water) 12.1 Normality HCl and 14.8 Normality NH4H2O to give 1M HCL and 1M NH4H20 respectively?

A)
12.1 N HCl contains 12.1 equivalents (or 12.1 mol) in 1 liter of water. To obtain 1 N HCl one would the volume equivalant to 1 mol HCl and dilute that volume to a total volume of 1 liter (or 1000 ml). This can be solved by a simple proportion 1 mol / 12.1 mol = x ml / 1000 ml, solving for x, x = 1000 ml x 1 mol / 12.1 mol = 82.64 ml.

So to obtain 1M HCl one would take 82.64 ml and dilute it to 1 liter (add 917.36 ml water)

B)
You state NH4H2O. This structure doesn't make sense. I believe you mean NH4OH. In any case, the same reasoning applies as in A) above.

1 mol / 14.8 mol = x ml / 1000 ml, solving for x, x = 1000 ml x 1 mol / 14.8 mol = 67.57 ml.

So to obtain 1M NH4OH one would take 67.57 ml and dilute it to 1 liter (add 932.43 ml water)

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One final point, question 2 refers to concentration in N (normality) rather then M (molarity). Normality is generally used for acids and bases to reflect the number of protons or hydroxyl anions that are liberated on ionization. In the case of HCl, 1 proton is given up and NH4OH, 1 hydroxyl is liberated on ionization. So in both cases the normalilty is equal to the molarity. For sulfuric acid = H2SO4, 2 protons are liberated on ionization and a 1 N solution is 0.5 M (since 1 mole contains 2 ionizable protons).

Hope this helps.

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