Question

The amount of acetic acid present in 100ml of 0.1M solution is:
A. 0.30g
B. 3.0g
C. 0.60g
D. None

Answer 1

Hint: To solve this problem, first we have to calculate the molar mass of acetic acid present in \[100ml\]of \[0.1M\]solution. Molar mass is the mass of all the atoms of a molecule in grams per mole. Then we will use the molarity formula to find out the amount of acetic acid present in solution.

Complete step by step solution:
First of all let us learn about the concepts provided in the solution and find out the correct solution for the given question.
The molar mass of a chemical compound is defined as the mass of a sample of that compound divided by the amount of substance in that sample, measured in moles.
The molarity (M) of a solution is the quantity of moles of solute dissolved in one liter of solution. To ascertain the molarity of a solution, you divide the moles of solute by the volume of the solution communicated in liters.
The formula of acetic acid is \[C{{H}_{3}}COOH\]
Adding the molecular mass of the elements to get the molar mass of acetic acid:
\[C{{H}_{3}}COOH\]
Molar mass \[=\left( 2\times 12.011\left( C \right)+4\times 1.00794\left( H \right)+2\times 15.999\left( O \right) \right)\]
\[=60g\]
\[{{M}_{acetic\; acid}}=60g\]
Molarity \[=\dfrac{mass}{{{M}_{mass}}}\times \dfrac{1000}{V\left( ml \right)}\]
 \[\Rightarrow 0.1=\dfrac{mass}{60}\times \dfrac{1000}{100}\]
 \[\Rightarrow m=0.60g\]

Hence, the correct option is C. \[0.60g\]

Additional Information: Acetic Acid, \[C{{H}_{3}}COOH\], is a week corrosive, since it is available in solution principally as whole CH3COOH atoms, and next to no as \[{{H}^{+}}\]and \[C{{H}_{3}}CO{{O}^{-}}\] particles. Which besides shows that acetic acid is frail, on the grounds that strong ions ionize totally.

Note: Molar concentration (likewise called molarity, amount concentration or substance concentration) is a proportion of the concentration of a chemical variety, specifically of a solute in a solution, regarding amount of substance per unit volume of solution.