Question

Calculate the mass of CH3COOH present in 200ml of 0.5 molar solution?

Answer 1

Answer:

6

Explanation:

0.5=(g.m×1000)/(m.m×vol in ml)

Answer 2

Concept:

Molarity: The molarity of a solution is defined as the number of moles of solute dissolved per unit volume of the solution. It is denoted by M and is given by,

M = n/V

where n = Number of moles of solute

V = Volume of the solution

Number of moles of the solute is given by,

n = m/W

where m = Mass of the solute in grams

W = Molecular mass of the solute

Given:

Molarity, M = 0.5 M

Volume of the solution, V = 200 mL = 0.2 L

Solute = Acetic acid = CH₃COOH

Find:

Mass of CH₃COOH present in 200 mL of 0.5 molar solution.

Solution:

First of all, we will calculate the molecular mass of the solute.

Molecular mass of CH₃COOH = Atomic mass of C + 3×Atomic mass of H + Atomic mass of C + Atomic mass of O + Atomic mass of O + Atomic mass of H

Molecular mass of CH₃COOH, W = 12 + 3(1) + 12 + 16 + 16 + 1

                                             W = 60 g mol⁻¹

Molarity, M = 0.5             [Given]

Let number of moles be n.

Volume, V = 0.2 L

As we know,

M = n/V

0.5  = n/0.2

n = 0.5×0.2

n = 0.1

Now, Number of moles, n = 0.1

Let the mass of CH₃COOH be m.

Number of moles, n = m/W

0.1 = m/60

m = 0.1×60

m = 6 g

∴ Mass of CH₃COOH, m = 6 g

Hence, the mass of CH₃COOH present in 200 mL of 0.5 molar solution is 6 grams.

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