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

Answer:

$m_{fresh}= 1000 \frac{Kg}{m^3} * 2.5x10^{-4} m^3 = 0.25 Kg$

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:

$m_{sea}= 1030 \frac{Kg}{m^3} * 1x10^{-4} m^3 = 0.103 Kg$

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

$\rho_{mixture}= \frac{m_{fresh} +m_{sea}}{v_{fresh} +v_{sea}}$

And replacing we got:

$\rho_{mixt}= \frac{0.25 +0.103 Kg}{2.5x10^{-4} m^3 +1x10^{-4} m^3} = 1008.571 \frac{Kg}{m^3}$

Step-by-step 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:

$m_{fresh}= 1000 \frac{Kg}{m^3} * 2.5x10^{-4} m^3 = 0.25 Kg$

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:

$m_{sea}= 1030 \frac{Kg}{m^3} * 1x10^{-4} m^3 = 0.103 Kg$

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

$\rho_{mixture}= \frac{m_{fresh} +m_{sea}}{v_{fresh} +v_{sea}}$

And replacing we got:

$\rho_{mixt}= \frac{0.25 +0.103 Kg}{2.5x10^{-4} m^3 +1x10^{-4} m^3} = 1008.571 \frac{Kg}{m^3}$