SSCE VCE Chemistry Families of Organic Compounds: 0.132 g of a pure carboxylic aci

0.132 g of a pure carboxylic acid (R-COOH) was dissolved in 25.00 mL of water and titrated with 0.120 M NaOH solution - VCE - SSCE Chemistry - Question 8 - 2009 - Paper 1

Question 8

0.132 g of a pure carboxylic acid (R-COOH) was dissolved in 25.00 mL of water and titrated with 0.120 M NaOH solution. A volume of 14.80 mL was required to reach the... show full transcript

Worked Solution & Example Answer:0.132 g of a pure carboxylic acid (R-COOH) was dissolved in 25.00 mL of water and titrated with 0.120 M NaOH solution - VCE - SSCE Chemistry - Question 8 - 2009 - Paper 1

Step 1

Calculate the number of moles of NaOH used

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Answer

To find the number of moles of NaOH used in the titration, we use the formula:

$n = C \times V$

where:

$n$ is the number of moles,
$C$ is the concentration (in mol/L),
$V$ is the volume (in L).

Using the given data: $C = 0.120 \text{ M}, \, V = 14.80 \text{ mL} = 0.01480 \text{ L}$ $n_{NaOH} = 0.120 \times 0.01480 = 0.001776 \text{ moles}$

Step 2

Determine the moles of the carboxylic acid

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Answer

In a neutralization reaction between a carboxylic acid and NaOH, the reaction ratio is 1:1. Therefore, the moles of the carboxylic acid (R-COOH) will be the same as those of NaOH: $n_{R-COOH} = n_{NaOH} = 0.001776 \text{ moles}$

Step 3

Calculate the molar mass of the carboxylic acid

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Answer

The molar mass (M) can be calculated using: $M = \frac{\text{mass}}{n}$ In this case:

mass = 0.132 g,
$n = 0.001776 \text{ moles}$ . Thus: $M = \frac{0.132}{0.001776} = 74.4 \text{ g/mol}$

Step 4

Identify the possible carboxylic acid

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Answer

Let's check the molar masses of the given options:

A. HCOOH: 46.03 g/mol (not correct)
B. CH₃COOH: 60.05 g/mol (not correct)
C. C₂H₅COOH: 74.08 g/mol (closest, but slightly higher)
D. C₃H₇COOH: 74.40 g/mol (correct)

Thus, the answer is D. C₃H₇COOH.

0.132 g of a pure carboxylic acid (R-COOH) was dissolved in 25.00 mL of water and titrated with 0.120 M NaOH solution - VCE - SSCE Chemistry - Question 8 - 2009 - Paper 1

Worked Solution & Example Answer:0.132 g of a pure carboxylic acid (R-COOH) was dissolved in 25.00 mL of water and titrated with 0.120 M NaOH solution - VCE - SSCE Chemistry - Question 8 - 2009 - Paper 1

Calculate the number of moles of NaOH used

Determine the moles of the carboxylic acid

Calculate the molar mass of the carboxylic acid

Identify the possible carboxylic acid

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