Two miscible liquids of densities $1.2 g m / c c$ and $1.4 g m / c c$ are mixed with a proportion of their volumes equal to $3 : 5$ . What is the density of resulting liquid?

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Solution

Let the volume of first liquid be $V_{1} = 3 x$
Volume of second liquid $V_{2} = 5 x$
Given : $ρ_{1} = 1.2 g / c m^{3}$ $ρ_{2} = 1.4 g m / c m^{3}$
Mass of first liquid $m_{1} = ρ_{1} V_{1} = 3.6 x$
Mass of second liquid $m_{2} = ρ_{2} V_{2} = 7 x$
So, Density $ρ = \frac{m_{1} + m_{2}}{V_{1} + V_{2}} = \frac{3.6 x + 7 x}{1.2 x + 1.4 x} = \frac{10.6}{2.6} = 4.07 g m / c m^{3}$

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Two miscible liquids of densities 1.2gm/cc and 1.4gm/cc are mixed with a proportion of their volumes equal to 3:5. What is the density of resulting liquid?

Let the volume of first liquid be V1=3xVolume of second liquid V2=5 xGiven : ρ1=1.2 g/cm3 ρ2=1.4 gm/cm3Mass of first liquid m1=ρ1V1=3.6 xMass of second liquid m2=ρ2V2=7 xSo, Density ρ=m1+m2V1+V2=3.6x+7x1.2x+1.4x=10.62.6=4.07 gm/cm3

Two miscible liquids of densities $1.2 g m / c c$ and $1.4 g m / c c$ are mixed with a proportion of their volumes equal to $3 : 5$ . What is the density of resulting liquid?