Question

250cm3 of fresh water of density 1000kgm-3 is mixed with 100cm3 of sea water of density 1030kgm-3. Calculate the density of the mixture. *​

Answer 1

`m_(fresh)= 1000 (Kg)/(m^3) * 2.5x10^(-4) m^3 = 0.25 Kg`

And we can do a similar procedure for the sea water:

`m_(sea)= \rho_(sea) V_(sea)`

And after convert the volume to m^3 we got:

`m_(sea)= 1030 (Kg)/(m^3) * 1x10^(-4) m^3 = 0.103 Kg`

And then the density for the mixture would be given by:

`\rho_(mixture)= (m_(fresh) +m_(sea))/(v_(fresh) +v_(sea))`

And replacing we got:

`\rho_(mixt)= (0.25 +0.103 Kg)/(2.5x10^(-4) m^3 +1x10^(-4) m^3) = 1008.571 (Kg)/(m^3)`

Explanation:

For this case we can begin calculating the mass for each type of water:

`m_(fresh)= \rho_(fresh) V_(fresh)`

And after convert the volume to m^3 we got:

`m_(fresh)= 1000 (Kg)/(m^3) * 2.5x10^(-4) m^3 = 0.25 Kg`

And we can do a similar procedure for the sea water:

`m_(sea)= \rho_(sea) V_(sea)`

And after convert the volume to m^3 we got:

`m_(sea)= 1030 (Kg)/(m^3) * 1x10^(-4) m^3 = 0.103 Kg`

And then the density for the mixture would be given by:

`\rho_(mixture)= (m_(fresh) +m_(sea))/(v_(fresh) +v_(sea))`

And replacing we got:

`\rho_(mixt)= (0.25 +0.103 Kg)/(2.5x10^(-4) m^3 +1x10^(-4) m^3) = 1008.571 (Kg)/(m^3)`