Las disoluciones comerciales de HCl (PF = 36.5 g/mol) típicamente son 39.0% P/P de este ácido en agua. Determine la molalidad de la disolución del HCl si la densidad de la misma es igual a 1.20 g/ml.

Answers

Answer 1

The translation of the question is

Commercial solutions of HCl (MP = 36.5 g / mol) are typically 39.0% W / W of this acid in water. Determine the molality of the HCl solution if its density is equal to 1.20 g / ml.

Answer:

Molality is 17.51 molal

Explanation:

We are given 39% w/w of HCl, this means for every 100g of this solution we have 39g of HCl in it.

to ease the calculation we assume we have 1L(1000ml) of this solution is present.

Density = mass /volume

1.20 = mass/1000

mass = 1200 g of solution

and we are given that 39% of this solution is will contain HCl

so

1200 * 39/100 = 468g of HCl present

no of moles of HCl = 468/36.5 = 12.82moles

mass of solution = mass of water + mass of acid

mass of water = mass of solution - mass of acid

Mass of water = 1200 - 468 = 732g of water

Molality = moles /masss of water in kg

             = 12.82 *1000/732  =17.51molal


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Answers

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To calculate the amount of grams present in a given amount of moles, the molar mass of the substance must be used, as follows:

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In this way, just perform the following operation:

                           

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Learn more about mole calculation in: brainly.com/question/15693285

2. Al(OH),(s) + 3 HCl(aq) à 3 H2O(l) + AlCl3(aq). This reaction shows how aluminum hydroxide
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I

Answers

This tablet insufficient to neutralize the acid in the patient's stomach

Further explanation

Reaction

Al(OH)₃(s) + 3HCl(aq) ⇒ 3 H₂O(l) + AlCl₃(aq)

0.25 g of  Aluminum hydroxide-Al(OH)₃ , mol (MW=78 g/mol) :

[tex]\tt \dfrac{0.25}{78}=0.00321[/tex]

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mass HCl (MW=36,46 g/mol) :

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Answers

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Answers

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Answers

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