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riadik2000 [5.3K]
2 years ago
15

A solution is prepared by dissolving 0.5636 g oxalic acid (H2C2O4) in enough water to make 100.0 mL of solution. A 10.00-mL aliq

uot (portion) of this solution is then diluted to a final volume of 250.0 mL. What is the final molarity of the diluted oxalic acid solution
Chemistry
1 answer:
zepelin [54]2 years ago
6 0

Answer:

The final molarity of the diluted oxalic acid solution is 0.002504 M

Explanation:

Step 1: Data given

Mass of oxalic acid (H2C2O4) = 0.5636 grams

Volume of the solution = 100.0 mL = 0.100 L

A 10 ml of this solution diluted in 250 ml of solution.

Molecular weight of H2C2O4  = 90.03 g/mol

Step 2: Calculate  initial moles of H2C2O4

Moles H2C2O4  = mass / molar mass

Moles H2C2O4  = 0.5636 grams / 90.03 g/mol

Moles H2C2O4 = 0.00626 moles

Step 3: Calculate molarity of the solution

Molarity = moles / volume

Molarity = 0.00626 moles / 0.100 L

Molarity = 0.0626 M

Step 4: Calculate moles of a 10.00 mL aliquot

Moles = 0.0626 M * 0.010 L

Moles = 0.000626 moles

Step 5: Calculate the new molarity

Molarity = 0.000626 moles / 0.250 L

Molarity = 0.002504 M

The final molarity of the diluted oxalic acid solution is 0.002504 M

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Explanation:

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                PV = nRT

or,          n = \frac{PV}{RT}

                = \frac{1 atm \times 4.41 ml}{0.0821 Latm/mol K \times 293 K}      (as 1 bar = 1 atm (approx))

                = 0.183 mol

As,   Density = \frac{mass}{volume}

Hence, mass of water will be as follows.

                Density = \frac{mass}{volume}

             0.998 g/ml = \frac{mass}{3.26 ml}    

                 mass = 3.25 g

Similarly, calculate the moles of water as follows.

        No. of moles = \frac{mass}{\text{molar mass}}

                              =  \frac{3.25 g}{18.02 g/mol}            

                              = 0.180 mol

Moles of hydrogen = 0.180 \times 2 = 0.36 mol

Now, mass of carbon will be as follows.

       No. of moles = \frac{mass}{\text{molar mass}}

          0.183 mol =  \frac{mass}{12 g/mol}            

                              = 2.19 g

Therefore, mass of oxygen will be as follows.

              Mass of O = mass of sample - (mass of C + mass of H)

                                = 3.50 g - (2.19 g + 0.36 g)

                                = 0.95 g

Therefore, moles of oxygen will be as follows.

          No. of moles = \frac{mass}{\text{molar mass}}

                               =  \frac{0.95 g}{16 g/mol}            

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                         C              H           O

No. of moles:  0.183        0.36       0.059

On dividing:      3.1           6.1            1

Therefore, empirical formula of the given compound is C_{3}H_{6}O.

Thus, we can conclude that empirical formula of the given compound is C_{3}H_{6}O.            

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3 years ago
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