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riadik2000 [5.3K]
3 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]3 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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Analysis of a gaseous chlorofluorocarbon, CClxFy, shows that it contains 11.79% C and 69.57% Cl. In another experiment, you find
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Answer:

The molecular formula = C_2Cl_{4}F_2

Explanation:

Moles =\frac {Given\ mass}{Molar\ mass}

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<u>% moles of C = 11.79 / 12.0107 = 0.9816</u>

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<u>% moles of Cl = 69.57 / 35.453 = 1.9623</u>

Given that the gaseous chlorofluorocarbon only contains chlorine, flourine and carbon. So,

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<u>% moles of F = 18.64 / 18.998 = 0.9812</u>

Taking the simplest ratio for C, Cl and F as:

0.9816 : 1.9623 : 0.9812

= 1 : 2 : 1

The empirical formula is = CCl_2F

Also, Given that:

Pressure = 21.3 mm Hg

Also, P (mm Hg) = P (atm) / 760

Pressure = 21.3 / 760 = 0.02803 atm

Temperature = 25 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (25 + 273.15) K = 298.15 K  

Volume = 458 mL  = 0.458 L (1 mL = 0.001 L)

Using ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

0.02803 atm × 0.458 L = n × 0.0821 L.atm/K.mol × 298.15 K  

⇒n = 0.00052445 moles

Given that :  

Amount  = 0.107 g  

Molar mass = ?

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

0.00052445= \frac{0.107\ g}{Molar\ mass}

Molar\ mass= 204.0233\ g/mol

Molecular formulas is the actual number of atoms of each element in the compound while empirical formulas is the simplest or reduced ratio of the elements in the compound.

Thus,  

Molecular mass = n × Empirical mass

Where, n is any positive number from 1, 2, 3...

Mass from the Empirical formula = 1×12.0107 + 2×35.453 + 1×18.998 = 101.9147 g/mol

Molar mass = 204.0233 g/mol

So,  

Molecular mass = n × Empirical mass

204.0233 = n × 101.9147

⇒ n = 2

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