Make the aqueous solutions, which contain 1 mole of solute per 100 mol of solven
ID: 477974 • Letter: M
Question
Make the aqueous solutions, which contain 1 mole of solute per 100 mol of solvent. Will need 300 mL of the acid solution and 600 mL of the NaOH. Assume that 1 L of water is 55.5 moles and add the necessary amount of solute to produce the 1 to 100 Molar ratio. SHOW ALL CALCULATIONS DEMONSTRATING HOW MUCH SOLUTE IS NEEDED VS. SOLVENT. Should have two answers, one for the NaOH solutions and one for the acetic acid solution. HCL: Concentrated HCl (molarity 12.1) is 36% by volume, with a molar mass of 36.46g/mol NaOH: has a molar mass of 40g/mol. CH3COOH: Concentrated acetic acid has a density of 1.049 g/mL with a molar mass of 60.05 g/mol. Assume it is pure acetic acid. H20: has a molarity of 55.5 and a density of 1.000g/mL
Explanation / Answer
Synthesis of acid solution:
1L of water = 55.5 moles
100 mol solvent = 1*100mol/55.5 mol =1.80 L
1.80 L solvent will contain 1 mole acetic acid.
moles of acetic acid present in 300mL = 1 mol *0.3L/1.8L = 0.167 mol
So, u have to add 0.167 mol acid in 300mL of solvent.
Mass of acetic acid used = 0.167 mol * 60.06 g/mol = 10 g
Now, as acetic acid is 100% pure, volume of 1 mL acetic acid = 1.049gm (from density value given)
So, volume of 10gm acetic acid = 1mL * 10gm/1.049 gm = 9.54 mL
Amount od acetic acid required to produce 1:100 ratio = 9.54 mL
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Calculations for NaOH.
1L of water = 55.5 moles
100 mol solvent = 1*100mol/55.5 mol =1.80 L
1.80 L solvent will contain 1 mole NaOH.
moles of NaOH present in 600mL = 1 mol *0.6L/1.8L = 0.333 mol
So, u have to add 0.333 mol base in 600mL of solvent.
Mass of acetic acid used = 0.333 mol * 40 g/mol = 13.2 g
13.2 gm NaOh has to be mixed with 600mL of water to get the desires ratio.
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