On domestic products such as vinegar, the concentration of active ingredients is rarely reported in Molar units. It is more common to report the concentration as a composition such as mass per 100 mL or volume per 100 mL. Vinegar is usually reported as a volume composition (volume of acetic acid per 100 mL vinegar - written as % v/v). To convert a molar concentration to % v/v:Calculate how many moles of acetic acid is in 100 mL of solution (Hint: n = C x v)Calculate the mass of acetic acid from the amount of moles determined in step 1 (Hint: m= MW x n)Calculate the volume that the mass determined in step 2 would occupy. To do this, you will have to divide the mass (determined in step 2) by the density of acetic acid (1.049 g.mL⁻¹). The density states how much 1 mL of a liquid sample weighs - for acetic acid, 1 mL weighs 1.049 g.The answer in step 3 states what volume of acetic acid is present in a 100 mL sample of vinegar which equates to the composition we are after. So we can just add "% v/v" to the end of our value to denote this is the volume of acetic acid per 100 mL of vinegar. If, from titration, you determined that concentration of acetic acid in vinegar is 0.928 M, what is the composition of acetic acid in vinegar (% v/v). Answer to 3 significant figures (please enter numeric value only)
Question
On domestic products such as vinegar, the concentration of active ingredients is rarely reported in Molar units. It is more common to report the concentration as a composition such as mass per 100 mL or volume per 100 mL. Vinegar is usually reported as a volume composition (volume of acetic acid per 100 mL vinegar - written as % v/v). To convert a molar concentration to % v/v:Calculate how many moles of acetic acid is in 100 mL of solution (Hint: n = C x v)Calculate the mass of acetic acid from the amount of moles determined in step 1 (Hint: m= MW x n)Calculate the volume that the mass determined in step 2 would occupy. To do this, you will have to divide the mass (determined in step 2) by the density of acetic acid (1.049 g.mL⁻¹). The density states how much 1 mL of a liquid sample weighs - for acetic acid, 1 mL weighs 1.049 g.The answer in step 3 states what volume of acetic acid is present in a 100 mL sample of vinegar which equates to the composition we are after. So we can just add "% v/v" to the end of our value to denote this is the volume of acetic acid per 100 mL of vinegar. If, from titration, you determined that concentration of acetic acid in vinegar is 0.928 M, what is the composition of acetic acid in vinegar (% v/v). Answer to 3 significant figures (please enter numeric value only)
Solution
Apologies for the confusion, but it seems there's a mix-up in the instructions. To answer the original question, here are the steps:
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Calculate the number of moles of acetic acid in 100 mL of solution. Given that the molar concentration (C) is 0.928 M and the volume (v) is 100 mL or 0.1 L, we use the formula n = C x v. So, n = 0.928 moles/L x 0.1 L = 0.0928 moles.
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Calculate the mass of acetic acid from the number of moles. The molar mass (MW) of acetic acid is approximately 60.052 g/mol. Using the formula m = MW x n, we get m = 60.052 g/mol x 0.0928 moles = 5.57 g.
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Calculate the volume that this mass would occupy. Given that the density of acetic acid is 1.049 g/mL, we divide the mass by the density to get the volume. So, volume = 5.57 g / 1.049 g/mL = 5.31 mL.
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The answer in step 3 is the volume of acetic acid present in a 100 mL sample of vinegar. To express this as a percentage (v/v), we divide the volume of acetic acid by the total volume and multiply by 100. So, % v/v = (5.31 mL / 100 mL) x 100 = 5.31 % v/v.
So, the composition of acetic acid in the vinegar is approximately 5.31 % v/v.
Similar Questions
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