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Masteriza [31]
3 years ago
6

Calculate the molality of a 24.4% (by mass) aqueous solution of phosphoric acid (h3po4).

Chemistry
1 answer:
user100 [1]3 years ago
8 0
Molality pertains to a concentration of a mixture expressed in moles of solute per kg of solvent. Based on the given problem, the solute is the phosphoric acid with molecular weight of 97.99 g/mol. Assuming 1 g of sample, the mole of solute is equal to 0.244*(97.99) = 2.49x10-3 moles. The remaining percentage,t that is, 1-0.244 is the amount of solvent, which is equal to 0.756 g based on 1 gram assumption. Therefore, the molality is,
(0.244)*(mol/97.99 g)/(1-0.244)*(1 kg/1000 g) = 3.29 molala
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Answer:

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Solution:

As the chemist needs to prepare 250 mL of solution from 6.00 M ammonium hydroxide solution to prepare a 2.50 M aqueous solution of ammonium hydroxide.

Now

first he have to determine the amount of ammonium hydroxide solution that will be taken from6.00 M ammonium hydroxide solution

For this Purpose we use the following formula

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Put values from given data in the formula

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Rearrange the equation

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                    V1 = 104 mL

So 104 mL is the volume of the solution which we have to take from the 6.00 M ammonium hydroxide solution to prepare 2.5 M  aqueous solution of ammonium hydroxide

But we have to prepare 250 mL of the solution.

so the chemist will take 104 mL from 6.00 M ammonium hydroxide solution and have to add 146 mL water to make 250 mL of new solution.

in this question you have to tell about the amount of water that is 146 mL

250 mL (total solution)  = 104 mL (stock solution)  + 146 mL (water)

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