Calculate the molarity and the molality of a solution prepared by dissolving 2.00 g of ethanol (C2H5OH) with 100.0 g of water to give a final volume that is 102 mL?

1 Answer
Meave60
Feb 7, 2018

The molarity is #"0.426 M"#.

The molality is #"0.434 m"#.

Explanation:

We will need the moles ethanol for both calculations.

#2.00color(red)cancel(color(black)("g C"_2"H"_5"OH"))xx(1"mol C"_2"H"_5"OH")/(46.069color(red)cancel(color(black)("g C"_2"H"_5"OH")))="0.043413 mol C"_2"H"_5"OH"#

I am leaving a couple of extra digits to reduce rounding errors. The final answers will be rounded to three significant figures.

MOLARITY

#"molarity"="moles solute"/"L of solution"#

We need to convert #"102 mL of solution"# to #"L"#.

#"1 L"##=##"1000 mL"#

#102color(red)cancel(color(black)("mL"))xx(1"L")/(1000color(red)cancel(color(black)("mL")))="0.102 L"#

#"molarity"=("0.043413 mol C"_2"H"_5"OH")/(0.102"L")="0.426 mol C"_2"H"_5"OH/L"="0.426 M"# (rounded to three significant figures)

MOLALITY

#"molality"="moles solute"/"kg solvent"#

Water is the solvent.

We need to convert #"100.0 g"# water to #"kg"#.

#"1 kg"##=##"1000 g"#

#100.0color(red)cancel(color(black)("g"))xx(1"kg")/(1000color(red)cancel(color(black)("g")))="0.1000 kg"#

#"molality"=(0.043413"mol C"_2"H"_5"OH")/(0.1000"kg H"_2"O")="0.434 mol/kg"="0.434 m"# (rounded to three significant figures)

Related questions
Impact of this question
3938 views around the world
You can reuse this answer
Creative Commons License