How many grams of the compound in the figure above are required to make 1 liter of a 0.5 M solution? (Note: The atomic masses, in daltons, are approximately 12 for carbon, 1 for hydrogen, and 16 for oxygen.)

This question is incomplete as there is no figure in the question but the compound in the figure is Ethanoic acid (CH3COOH). Hence, we can calculate the mass.

The answer is: 30grams

Explanation:

According to the question, the volume of the compound (CH3COOH) is 1litre while the molar concentration is 0.5M

Molarity or Molar concentration is calculated using the formula:

M= n/V

Where; M= Molarity (M) = 0.5M

n= number of moles (mol)

V= Volume (V) = 1L

Hence, number of moles (n) = M × V

n = 0.5 × 1

n = 0.5 moles.

To calculate the mass in grams of CH3COOH, we say;

Number of moles (n) × molar mass (MM)

Since atomic mass for (C= 12, H= 1 and O=16)

Molar mass of CH3COOH:

= 12 + 1(3) + 12 + 16 + 16 + 1

Molar mass of CH3COOH = 60g/mol

Mass (in grams) of CH3COOH = 0.5 moles × 60g/mol

Mass = 30grams.

Therefore, 30grams of CH3COOH compound are required to make 1 liter of a 0.5 M solution.