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2 years ago

Concentrated hydrochloric acid is 36% (Weight/Volume) hydrochloric acid and has a density of 1.18 g.cm -3.

Chemistry

1 answer:

Juli2301 [7.4K]2 years ago

6 0

i. The new concentration is 0.116 M

ii. The new concentration is 0.029 M

<h3>What is the concentration of the original hydrochloric acid acid?</h3>

The concentration of the original or stock hydrochloric acid is calculated as follows:

Molarity = Percentage concentration * Density * 1000/Molar mass * 100

Molarity of stock HCl = 36 * 1.18* 1000 /36.5 * 100

Molarity of stock HCl = 11.6 mol/dm

i. Using the dilution formula: M₁V₁ = M₂V₂

M₂ = M₁V₁ /V₂

M₂ = 11.6 * 10/1000

M₂ = 0.116 M

The new concentration = 0.116 M

ii. Using M₁V₁ = M₂V₂

M₂ = M₁V₁ /V₂

M₂ = 0.116 * 5/20

M₂ = 0.029 M

The new concentration = 0.029 M

In conclusion, the new concentrations are found using the dilution formula.

Learn more about molarity at: brainly.com/question/17138838

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Calculate the solubility of hydrogen in water at an atmospheric pressure of 0.380 atm (a typical value at high altitude).

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The question is incomplete, here is the complete question:

Calculate the solubility of hydrogen in water at an atmospheric pressure of 0.380 atm (a typical value at high altitude).

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<u>Explanation:</u>

To calculate the partial pressure of hydrogen gas, we use the equation given by Raoult's law, which is:

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$p_A$ = partial pressure of hydrogen gas = ?

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where,

$K_H$ = Henry's constant = $7.80\times 10^{-4}mol/L.atm$

$p_{H_2}$ = partial pressure of hydrogen gas = $1.9\times 10^{-7}atm$

Putting values in above equation, we get:

$C_{H_2}=7.80\times 10^{-4}mol/L.atm\times 1.9\times 10^{-7}atm\\\\C_{CO_2}=1.48\times 10^{-10}M$

Hence, the solubility of hydrogen gas in water at given atmospheric pressure is $1.48\times 10^{-10}M$

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