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Shalnov [3]
3 years ago
5

Suppose you are working with a NaOH stock solution but you need a solution with a lower concentration for your experiment. Calcu

late the volume of the 1.224 M stock NaOH solution needed to prepare 250.0 mL of 0.1281 M dilute NaOH solution.
Chemistry
1 answer:
Monica [59]3 years ago
7 0

Answer: The volume of the 1.224 M NaOH solution needed is 26.16 mL

Explanation:

In order to prepare the dilute NaOH solution, solvent is added to a given amount of the NaOH stock solution up to a final volume of 250.0 mL.

Since only solvent is added, the amount of the solute, NaOH, in the dilute solution is the same as in the volume taken from the stock solution.

Molarity (<em>M)</em> is calculated from the following equation:

<em>M</em> = <em>n</em> ÷ <em>V</em>

where <em>n</em> is the number of moles of the solute in the solution, and <em>V</em> is the volume of the solution.

Accordingly, the number of moles of the solute is given by

<em>n</em> = <em>M</em> x <em>V</em>

Now, let's designate the stock NaOH solution and the dilute solution as (1) and (2), respectively . The number of moles of NaOH in each of these solutions is:

<em>n </em>(1) = <em>M </em>(1) x <em>V </em>(1)

<em>n </em>(2) = <em>M </em>(2) x <em>V </em>(2)

As the amount of NaOH in the dilute solution is the same as in the volume taken from the stock solution,

<em>n</em> (1) = <em>n</em> (2)

and

<em>M</em> (1) x <em>V</em> (1)<em> </em>= <em>M</em> (2) x <em>V</em> (2)

For the stock solution, <em>M</em> (1) = 1.244 M, and <em>V</em> (1) is the volume needed. For the dilute solution, <em>M</em> (2) = 0,1281 M, and <em>V</em> (2) = 250.0 mL.

The volume of the stock solution needed, <em>V</em> (1), is calculated as follows:

<em>V</em> (1) = <em>M</em> (2) x <em>V</em> (2) ÷ <em>M</em> (1)

<em>V</em> (1) = 0.1281 M x 250.0 mL ÷ 1.224 M

<em>V </em>(1) = 26.16 mL

The volume of the 1.224 M NaOH solution needed is 26.16 mL.

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Answer:

Original Mass before 5 days = 82.97 gm

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where,

Nt = Mass Remains

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Equation (2) implies:

⇒ 65.1 = No (1/2) ^ (5/14.3)

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We have the following data:

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X(C₄H₄S) = moles C₄H₄S/ total moles

To calculate the moles, we need the molecular weight (MW) of each compound:

MW(C₄H₄S) = (4 x 12 g/mol) + (4 x 1 g/mol) + 32 g/mol = 84 g/mol

MW(C₇H₈) =  (7 x 12 g/mol) + (8 x 1 g/mol) = 92 g/mol

Thus, we calculate the moles of C₄H₄S by dividing the mass into the MW(C₄H₄S):

moles C₄H₄S = mass(C₄H₄S)/MW(C₄H₄S)= 8.10 g/(84 g/mol) = 0.096 moles

Then, we have to calculate the moles of C₇H₈. First, we need the mass, obtained from the product of the density by the volume:

mass(C₇H₈)= d(C₇H₈) x V(C₇H₈) = 0.867 g/mL x 250.0 mL = 216.75 g

Thus, we divide the mass of C₇H₈ into the MW to calculate the moles of C₇H₈:

moles C₇H₈ = mass(C₇H₈)/MW(C₇H₈) = 216.75 g/(92 g/mol) = 2.35 moles

The total moles is obtained from the addition of the moles of the solute (C₄H₄S) and the solvent (C₇H₈):

total moles = moles C₄H₄S + moles C₇H₈ = 0.096 moles + 2.35 moles = 2.45 moles

Finally, we calculate the mole fraction of C₄H₄S:

X(C₄H₄S) = moles C₄H₄S/ total moles = 0.096 moles/2.45 moles = 0.039

(b) The <u>molarity of C₄H₄S</u> is calculated as follows:

M(C₄H₄S) = moles C₄H₄S/1 liter solution

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M(C₄H₄S) = 0.096 moles/(0.250 L) = 0.384 mol/M = 0.384 M

(c) <u>Assuming that the volumes of solute and solvent are additive</u>, we can add the volumes of C₄H₄S and C₇H₈. First, we need the volume of C₄H₄S, which can be calculated from the mass and density:

V(C₄H₄S) = mass(C₄H₄S)/ d(C₄H₄S) = 8.10 g/(1.065 g/mL) = 7.606 mL = 0.0076 L

Now, we add the volumes:

total volume = V(C₇H₈) + V(C₄H₄S) = 0.250 L + 0.0076 L = 0.2576 L

Finally, we recalculate the <u>molarity of C₄H₄S</u>:

M(C₄H₄S)= moles C₄H₄S/ total volume = 0.096 moles/0.2576 L = 0.373 M

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