5. Calculate the number of moles and grams of methanol, CH3OH, that must be adde

Question

5. Calculate the number of moles and grams of methanol, CH3OH, that must be added to 2.00 moles of water to make a solution that has equal numbers of methanol and water molecules. How many total molecule does the resulting solution contain? 6. Vinegar is a 5.0% acetic acid solution, CH3COOH(aq), by mass. What is the molarity of vinegar? You may assume that the density of vinegar is 1.00 g/mL. 7. How many moles of 0.100 M NaOH are required to completely react with 1.00 L of the vinegar in Question 5?

Explanation / Answer

5.

Moles of water = 2.00 mol

1 mol of any substance contains 6.023 x 10^23 molecules.

No.of molecule of water = 2.00 mol x (6.023 x 10^23 molecules / 1.00 mol) = 12.046 x 10^24 molecules.

It was said that the solution contains equal numbers of methanol and water molecules.

So, number of methanol molecules = 12.046 x 10^24 molecules.

Thus, number of moles of methanol = 2.00 mol

molarmass of methanol = 32 g/mol

2.00 mol of methanol = 64.00 g of methanol

Thus,

number of moles of methanol = 2.00 mol

grams of methanol = 64.00 g

Total number of molecules = 12.046 x 10^24 molecules of water + 12.046 x 10^24 molecules

= 2.4092 x 10^24 molecules.

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6.

Given, acetic acid soltuion = 5.0% w/w

That is 5 g of acetic acid is present in 100 g of solution

Moles of acetic acid = mass / Molar mass = 5g / 60g/mol = 0.083 mol

Volume of the solution = Mass of solution / density = 100 g / 1.00g/ml = 100 mL = 0.1 L

Molarity = moles of acetic acid / Volume (in L) = 0.083 / 0.1 L = 0.83 M

Thus, Molarity of vinegar = 0.83 M

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7.

Molarity of vinegar from above question = 0.83 M

That means 0.83 moles of acetic acid is present in 1.0 L of solution.

The reaction between acetic acid and NaOH can be depicted as a chemical reaction.

CH3COOH + NaOH ---------> CH3COONa + H2O

The reaction shows that acetic acid and NaOH react in equimolar ratio.

M1V1 = M2V2

(0.83 M) ( 1.00L) = (0.100 M)(V2)

V2 = 8.3 L

Thus, 8.3 L of 0.100 M NaOH (0.83 mol) are required to completely react with 1.00 L of vinegar.

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