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To solve the problem given in the question we need to know the densities of the both the liquids, which have not been specified. Another problem is that while it is clear that one of the liquid is water, the identity of the other liquid, described as "meth" is not clear. I assume meth refers to methyl alcohol, also called methanol and will sole the problem accordingly.
From the standard available from other sources we know:
Density of water = 1 g/cm^3
Density of methanol = 0.7913 g/cm^3
Volume of water = 200 cm^3
Volume of methanol = 100 cm^3
Weight of a given volume of any substance can be calculated as:
Weight = Volume*Density
Weight of water = (Volume of water)*(Density of water)
= 200*1 = 200 g
Weight of methanol = (Volume of methanol*(Density of methanol)
= 100*0.7913 = 79.13 g
Total mass of the mixture
= Mass of water + Mass of methanol = 200 + 79.13 = 279.13 g
Total volume of the mixture
= Volume of water + Volume of methanol = 200 + 100 = 300 cm^3
Average density of the mixture
= (Mass of mixture)/(Volume of mixture) = 279.13/300 = 0.930433 g/cm^3
Since the two liquids, water and meth are mixed, we pressume they are mixtures and not dissolving and so they keep their identities separate, the higher density of the two go below and the lower density liquid stay in upper level, and the volume of the two is together is the sum of their separate volumes )as there is no dissolving),i.e (0.200+100)cm^3 = (0.02+0.01)m^3 = 0.03 m^3
If d1 and d2 are the densities of water and meth respectively, then the mass of water = volume of water*d1 =0.02d1 kg. The mass of meth = volume of meth*density of meth = 0.01*d1kg.
Therefore, the total mass of the mix = ( 0.02d1+0.01d2) kg
Therefore, the averge density of the mix = Total mass / Total volume = (0.02d1+0.01d1)/(0.03) = 0.6667d1+0.3333d2 kg/m^3.
If we substitute the individual densities of the liquids, we get the values for masses of the liquids and their average density.
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