Question

Hydrochloric acid is sold
commercially as 12.0 M solution.
How may moles of HCL are in
3000 mL of 12.0 M solution?
​

Answer 1

Remember, 2.28-molal means 2.28 moles of acetone in 1.00 kilogram of ethanol.

1) Determine volumes of acetone and ethanol, then total volume:

acetone

2.28 mol x 58.0794 g/mol = 132.421 g

132.421 g divided by 0.788 g/cm3 = 168.047 cm3

ethanol

1000 g divided by 0.789 g/cm3 = 1267.427 cm3

total volume

168.047 + 1267.427 = 1435.474 cm3

2) Determine molarity:

2.28 mol / 1.435 L = 1.59 M

Solution for mole fraction:

1) Determine moles of ethanol:

1000 g / 46.0684 g/mol = 21.71 mol

2) Determine mole fraction of acetone:

2.28 / (2.28 + 21.71) = 0.0950

Problem #12: Calculate the normality of a 4.0 molal sulfuric acid solution with a density of 1.2 g/mL.

Reminders:

N = #equivalents / L solution

#equivalents = molecular weight / n (n = number of H+ or OH¯ released per dissociation.)

molal = moles solute / kg solvent

Solution:

1) Determine grams of H2SO4 present:

4.0 molal = 4.0 moles H2SO4 / 1000 g solution

4.0 mol times 98.09 g/mol = 392.32 g

2) Determine equivalent weight for H2SO4:

98.09 g/mol / 2 dissociable hydrogen/mol = 49.05 g/equivalent

3) Determine # equivalents in 392.32 g:

392.32 g times (1 equivalent / 49.05 g) = 8.0 equivalents

4) Determine volume of soluti