Question

PLEASE HELP ASAP.

Ethyl alcohol is to be mixed with water to obtain a mixture with density 900 kg/m3. In what proportion by (a) weight and (b) volume must they be mixed?

Answer 1

First of all, let's write the density of water (W) and the density of alcohol (A):

`d_W = 1000 kg/m^3`

`d_A = 789 kg/m^3`
And the relationship between density d, mass m and volume V:

`d= (m)/(V) (1)`
The density of the mixture water+alcohol is 900 kg/m^3. The mass of this mixture is the sum of the masses of water and alcohol, and the volume of this mixture is the sum of the volumes of water and alcohol. So we can write

`d_T = 900 kg/m^3 = (m_W +m_A)/(V_W+V_A)` (2)
Now let's solve the two parts of the problem.

a) Let's rewrite (2) by replacing the volumes with
`V= (m)/(d)`:

`900 kg/m^3 = (m_W+m_A)/( (m_W)/(1000 )+ (m_A)/(789) )`
By solving this equation, we find:

`(m_A)/(m_W)= (0.1)/(0.14)=0.71`
This means that there are 0.71 kg of alcohol per each kg of water in the mixture.

b) Similarly, let's rewrite (2) by replacing the masses with
`m=dV`:

`900 kg/m^3 = (1000 V_W+789 V_A)/(V_W+V_A)`
Re-arranging and solving, we find

`(V_A)/(V_W)= (100)/(111)=0.90`
This means that there are 0.90 L of alcohol per each liter of water in the mixture.