# Pure water and alcohol are mixed to get a 60% (weight) alcohol solution. The densities in Kg/m3 of water, alcohol and the solution may be taken as 998, 798 and 895 respectively at 293 K. Estimate the molarity and molality??

Aug 17, 2017

$\text{molarity} = 11.6$ $\text{mol/L}$

$\text{molality} = 32.6$ $\text{mol/kg}$

#### Explanation:

of a solution, given the percent by mass of the solution, and density values.

• Let's do the molality first

Equation for molality:

$\text{molality" = "mol solute"/"kg solvent}$

We're given that a solution of alcohol and water is 60% alcohol by mass, so let's assume we have a $100$-$\text{g}$ sample, so we would then have

• $60$ $\text{g alcohol}$

• $40$ $\text{g water}$

(The "alcohol" here is ethanol, which isn't given to us, but we can either assume this, or look it up based on the given density.)

Let's convert from grams of ethanol to moles, using its molar mass:

60cancel("g ethanol")((1color(white)(l)"mol ethanol")/(46.068cancel("g ethanol"))) = color(red)(ul(1.30color(white)(l)"mol ethanol"

And the number of kilograms of water is

40cancel("g H"_2"O")((1color(white)(l)"kg H"_2"O")/(10^3cancel("g H"_2"O"))) = color(green)(ul(0.040color(white)(l)"kg H"_2"O"

The molality is thus

color(blue)("molality") = color(red)(1.30color(white)(l)"mol ethanol")/color(green)(0.040color(white)(l)"kg H"_2"O") = color(blue)(ulbar(|stackrel(" ")(" "32.6color(white)(l)"mol/kg"" ")|)

$\text{ }$

• Now, the molarity

The equation for molarity is

$\text{molarity" = "mol solute"/"L solution}$

To find the volume of solution, we take the given density, and the fact that we assumed a $100$-$\text{g}$ sample of solution:

$\text{desnity" = "mass"/"volume}$

So

$\text{volume" = "mass"/"density}$

The density of the solution is given as $895$ ${\text{kg/m}}^{3}$, so we have

100cancel("g soln")((1cancel("kg soln"))/(10^3cancel("g soln")))((1cancel("m"^3))/(895cancel("kg soln")))((100^3cancel("cm"^3))/(1cancel("m"^3)))((1cancel("mL soln"))/(1cancel("cm"^3)))((1color(white)(l)"L soln")/(10^3cancel("mL soln"))) = color(red)(ul(0.112color(white)(l)"L solution"

The molarity is therefore

color(darkblue)("molarity") = (1.30color(white)(l)"mol ethanol")/(color(red)(0.112color(white)(l)"L soln")) = color(darkblue)(ulbar(|stackrel(" ")(" "11.6color(white)(l)"mol/L"" ")|)