mass that ethylene glycol = #\"40 g\"#mass of water (solvent) = #\"60 g = 0.060 kg\"#mass of solution = #\"100 g\"#

1. Molality

Molar fixed of ethylene glycol = #\"62.07 g/mL\"#

no. Of mole = #\"4 g\"/(\"62.07 g.mol\"^-1) = \"0.6444 mol\"#

Molality = #\"no. Of moles\"/\"mass of solvent in kg\"#

Molality = #\"0.6444 mol\"/\"0.060 kg\" = \"10.7 m\"#

2. Molarity

#\"Molarity\" = \"no. The moles\"/\"volume of solution in litres\"#

#\"Density\" = \"1.05 g/cm\"^3#

#\"Volume\" = \"Mass\"/\"Density\"#

#\"Volume\" = \"100 g\"/(\"1.05 g/mL\") = \"95.24 mL\" = \"0.0952 L\"#

#\"Molarity\" =\"0.6444 mol\"/\"0.0952 L\" = \"6.76 M or 6.76 mol/L\"#

3. Mole fraction

#\"no. Of mole of water\" = \"60 g\"/\"18.02 g.mol\"^\"-1\" = \"3.329 mol\"#

#\"Total moles\" = \"(3.329 + 0.6444) mol\" = \"3.974 mol\"#

#\"mole portion of ethylene glycol\" = \"0.6444\"/\"3.974\" = 0.162\"#


Answer link
\"*\"

Truong-Son N.
Feb 16, 2017

#<\"C\"_2\"H\"_6\"O\"_2> = \"6.83 M\"#

#m_(\"C\"_2\"H\"_6\"O\"_2) = \"10.741 mol/kg\"#

#chi_(\"C\"_2\"H\"_6\"O\"_2) = 0.1621#

Read further to see how it was done.

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This is just an practice in flexing the borders of what girlfriend have and calculating various types of concentrations.

The solution is aqueous, for this reason the solvent is water, i m sorry is why the density is close to #\"1 g/mL\"#. Understanding the percent through mass, i m sorry is:

#\"% w/w\" = \"mass solute\"/\"mass solution\"xx100%#

we can assume #\"1000 g\"# solvent because that convenience (given that the molality is per #\"kg\"# of solvent) come get:

#\"40% w/w\" => 0.40 = \"x g solute\"/\"(1000 + x) g solution\"#

Solving because that #x#, us can get the fixed of the solute:

#0.40(1000 + x) = x#

#=> 400 + 0.40x = x#

#=> 400 = (1 - 0.40)x#

#=> x = 400/(1 - 0.40) = 666.bar(66)# #\"g solute\"#

Therefore, we can acquire the mols the solute and mols that solvent:

#color(green)(n_\"solute\") = (666.bar(66) \"g solute\")/(2xx12.011 + 6xx1.0079 + 2xx15.99\"9 g/mol\") = color(green)(\"10.741 mols ethylene glycol\")#

#color(green)(n_\"solvent\") = (\"1000 g solvent\")/(\"18.015 g/mol\") = color(green)(\"55.509 mols water\")#

From there, we have all the details we need to calculate the concentrations.

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MOLARITY

For the molarity:

#color(blue)(<\"C\"_2\"H\"_6\"O\"_2>) = \"mols solute\"/\"L solution\"#

#= \"10.741 mols solute\"/((1000 + 666.bar(66)) cancel\"g solution\" xx cancel\"mL\"/(1.06 cancel\"g\") xx \"L\"/(1000 cancel\"mL\"))#

#=# #color(blue)(\"6.83 M\")#

MOLALITY

The molality to be made basic because we made decision the fixed of the solvent to it is in #\"1000 g\"#, i.e. #\"1 kg\"#:

#color(blue)(m_(\"C\"_2\"H\"_6\"O\"_2)) = \"mols solute\"/\"kg solvent\"#

#= \"10.741 mols solute\"/\"1 kg water\"#

#=# #color(blue)(\"10.741 mol/kg\")#

Naturally, we determined #\"1000 g\"# that water so that we couldn\"t mess up this calculate as lengthy as we gained the mols appropriate (dividing by 1 is simple to acquire right).