Anonymous
Anonymous asked in Science & MathematicsChemistry · 1 month ago

# What is the molality of an ethanol-water solution if 64.3mL of ethanol(CH3CH2OH, d = 0.789 g/cm3) is added to 25.9g of water(d = 1.00 g/mL)?

A.

53.2 m

B.

0.0425 m

C.

36.2 m

D.

42.5 m

E.

67.4 m

Relevance
• What is the molality of an ethanol-water solution if 64.3mL of ethanol(CH3CH2OH, d = 0.789 g/cm3) is added to 25.9g of water(d = 1.00 g/mL)?

Density = Mass/ Volume

Mass of water=

Volume of water× density

=64.3×0.789

= 81.50 gm

Molar mass of ethanol/ Weight

46/0.0259=

1776.0617760617

Apply about value in formula of molality= wt of solute / weight of solvent

• molality = moles solute / kg solvent

in this particular case, I'd argue the ethanol was the solvent.  There's 2.5x as much EtOH as H2O

so that

.. 25.9g H2O... .. 1 mol H2O.. .. 1 mL EtOH. .. .1000g

---- ---- ---- ----- x ----- ----- ----- x ----- ----- ----- x ---- ----- = 28.3m

. 64.3mL EtOH.. 18.02g H2O.. .0.789g EtOH.. ..1 kg

********

but.. that's not one of the choices is it?  so let's see what happens if we...  for God only knows why... assume H2O is the solvent

.. 64.3mL EtOH.. . 0.789g EtOH.... 1 mol EtOH... .. 1000g

---- ----- ---- ----- x ---- ---- ---- ---- x ---- ---- ---- ---- x ---- ----- = 42.5m

.. .25.9g H2O.. .. ... .1 mL EtOH.. . 46.07g EtOH.. . .. 1 kg

• (64.3 mL) x (0.789 g/mL) / (46.0684 g ethanol/mol) / (0.0259 kg) = 42.5 mol / kg =

42.5 m