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bija089 [108]
3 years ago
12

A chemist makes 620 mL of barium chloride working solution by adding distilled water to180mL of a 1.39mol/L stock solution of ba

rium chloride in water. Calculate the concentration of the chemist's working solution. Be sure your answer has the correct number of significant digits.
Chemistry
2 answers:
katen-ka-za [31]3 years ago
7 0

Answer:

The correct answer is 0.40 mol/L.

Explanation:

In order to solve the problem, we use the following expression, where V1 and C1 are the volume and concentration of stock solution and V2 and C2 are the volume and concentration of working solution:

V1 x C1= V2 x C2

⇒C2= V1 x C1/ v2

  C2= (180 ml x 1.39 mol/L)/ 620 ml

  C2= 0.40 mol/L

Svetllana [295]3 years ago
6 0

Answer:

0.404 mol / L

Explanation:

The <em>total moles of barium chloride</em> can be calculated from <em>the volume and concentration of the stock solution</em>:

1.39 mol/L * 0.180 L = 0.2502 moles barium chloride

Now we <u>calculate the concentration of the working solution, by dividing the total moles by the final volume</u>:

0.2502 mol / 0.620 L = 0.404 mol / L

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Read 2 more answers
Question List (4 items) (Drag and drop into the appropriate area) Find the volume of HCl that will neutralize the base. Find the
expeople1 [14]

The question is incomplete, the complete question is:

The solubility of slaked lime, Ca(OH)_2, in water is 0.185 g/100 ml. You will need to calculate the volume of 2.50\times 10^{-3}M HCl needed to neutralize 14.5 mL of a saturated

<u>Answer:</u> The volume of HCl required is 290mL, the mass of Ca(OH)_2 is 0.0268g, the moles of

<u>Explanation:</u>

Given values:

Solubility of Ca(OH)_2 = 0.185 g/100 mL

Volume of Ca(OH)_2 = 14.5 mL

Using unitary method:

In 100 mL, the mass of Ca(OH)_2 present is 0.185 g

So, in 14.5mL. the mass of Ca(OH)_2 present will be =\frac{0.185}{100}\times 14.5=0.0268g

The number of moles is defined as the ratio of the mass of a substance to its molar mass.

The equation used is:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}} ......(1)

Given mass of Ca(OH)_2 = 0.0268 g

Molar mass of Ca(OH)_2 = 74 g/mol

Plugging values in equation 1:

\text{Moles of }Ca(OH)_2=\frac{0.0268g}{74g/mol}=0.000362 mol

Moles of OH^- present = (2\times 0.000362)=0.000724mol

The chemical equation for the neutralization of calcium hydroxide and HCl follows:

Ca(OH)_2+2HCl\rightarrow CaCl_2+2H_2O

By the stoichiometry of the reaction:

Moles of OH^- = Moles of H^+ = 0.000724 mol

The formula used to calculate molarity:

\text{Molarity of solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (mL)}} .....(2)

Moles of HCl = 0.000724 mol

Molarity of HCl = 2.50\times 10^{-3}

Putting values in equation 2, we get:

2.50\times 10^{-3}mol=\frac{0.000724\times 1000}{\text{Volume of solution}}\\\\\text{Volume of solution}=\frac{0.000725\times 1000}{2.50\times 10^{-3}}=290mL

Hence, the volume of HCl required is 290mL, the mass of Ca(OH)_2 is 0.0268g, the moles of

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