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neonofarm [45]
1 year ago
7

Concentrated HCl is 15 M (15 molar). A 500 mL quantity of this solution is diluted by adding water to give a final volume of 2.5

L. What is the molarity of the resulting solution
Chemistry
1 answer:
Murrr4er [49]1 year ago
7 0

The molarity of the resulting solution obtained by diluting the stock solution is 3 M

<h3>Data obtained from the question </h3>
  • Molarity of stock solution (M₁) = 15 M
  • Volume of stock solution (V₁) = 500 mL
  • Volume of diluted solution (V₂) = 2.5 L = 2.5 × 1000 = 2500 mL
  • Molarity of diluted solution (M₂) =?

<h3>How to determine the molarity of diluted solution </h3>

M₁V₁ = M₂V₂

15 × 500 = M₂ × 2500

7500 = M₂ × 2500

Divide both side by 2500

M₂ = 7500 / 2500

M₂ = 3 M

Thus, the volume of the resulting solution is 3 M

Learn more about dilution:

brainly.com/question/15022582

#SPJ1

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b. Its concentration is half that of the chloride ion

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A student reacts 5.0 g of sodium with 10.0 g of chlorine and collect 5.24 g of sodium chloride. What is the percent yield of thi
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Answer: The percent yield of this combination reaction is 41.3 %

Explanation : Given,

Mass of Na = 5.0 g

Mass of Cl_2 = 10.0 g

Molar mass of Na = 23 g/mol

Molar mass of Cl_2 = 71 g/mol

First we have to calculate the moles of Na and Cl_2.

\text{Moles of }Na=\frac{\text{Given mass }Na}{\text{Molar mass }Na}

\text{Moles of }Na=\frac{5.0g}{23g/mol}=0.217mol

and,

\text{Moles of }Cl_2=\frac{\text{Given mass }Cl_2}{\text{Molar mass }Cl_2}

\text{Moles of }Cl_2=\frac{10.0g}{71g/mol}=0.141mol

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation will be:

2Na+Cl_2\rightarrow 2NaCl

From the balanced reaction we conclude that

As, 2 mole of Na react with 1 mole of Cl_2

So, 0.217 moles of Na react with \frac{0.217}{2}=0.108 moles of Cl_2

From this we conclude that, Cl_2 is an excess reagent because the given moles are greater than the required moles and Na is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of NaCl

From the reaction, we conclude that

As, 2 mole of Na react to give 2 mole of NaCl

So, 0.217 mole of HCl react to give 0.217 mole of NaCl

Now we have to calculate the mass of NaCl

\text{ Mass of }NaCl=\text{ Moles of }NaCl\times \text{ Molar mass of }NaCl

Molar mass of NaCl = 58.5 g/mole

\text{ Mass of }NaCl=(0.217moles)\times (58.5g/mole)=12.7g

Now we have to calculate the percent yield of this reaction.

Percent yield = \frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100

Actual yield = 5.24 g

Theoretical yield = 12.7 g

Percent yield = \frac{5.24g}{12.7g}\times 100

Percent yield = 41.3 %

Therefore, the percent yield of this combination reaction is 41.3 %

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What volume is used to calculate the volume of a solid object ​
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