Physics, 04.11.2020 18:30 asmita9876

Liquid A has a mass density of 850.0 kg/m3 and Liquid B has a mass density of 1060.0 kg/m3. Seventy-five grams of each liquid is mixed uniformly. What is the specific gravity of the mixture

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Answer from: avasteinhebel10

The specific gravity of the mixture of the liquid A and liquid B is 0.8.

The given parameters;

density of liquid A, = 850 kg/m³density of liquid B, = 1060 kg/m³mass of each liquid, m = 75 g = 0.075 kg

The volume of each liquid is calculated as follows;

$volume = \frac{mass}{density} \\\\volume_{A} = \frac{0.075}{850} \\\\volume_{A} = 8.82 \times 10^{-5}\\\\volume_{B} = \frac{0.075}{1060} \\\\volume_{B} = 7.08 \times 10^{-5}$

The specific gravity of the mixture is calculated as;

$S.G = \frac{\rho _A}{\rho_B} \\\\S.G = \frac{m/V_A}{m/V_B} \\\\S.G = \frac{V_B}{V_A} \\\\S.G = \frac{7.08 \times 10^{-5}}{8.82 \times 10^{-5} } \\\\S.G = 0.8$

Thus, the specific gravity of the mixture of the liquid A and liquid B is 0.8.

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Answer from: makennahudson94

The specific gravity of the mixture is 0.943.

Explanation:

The density of the mixture ( $\rho_{M}$ ), measured in kilograms per cubic meters, is defined by the following equation:

$\rho_{M} = \frac{m_{A}+m_{B}}{V_{A}+V_{B}}$ (Eq. 1)

Where:

$m_{A}$ , $m_{B}$ - Masses of liquids A and B, measured in kilograms.

$V_{A}$ , $V_{B}$ - Volumes of liquids A and B, measured in cubic meters.

By applying the definition of density, we expand the density of the mixture as follows:

$\rho_{M} = \frac{m_{A}+m_{B}}{\frac{m_{A}}{\rho_{A}}+\frac{m_{B}}{\rho_{B}} }$

$\rho_{M} = \frac{m_{A}+m_{B}}{\frac{\rho_{B}\cdot m_{A}+\rho_{A}\cdot m_{B}}{\rho_{A}\cdot \rho_{B}} }$

$\rho_{M} = \frac{\rho_{A}\cdot \rho_{B}\cdot (m_{A}+m_{B})}{\rho_{B}\cdot m_{A}+\rho_{A}\cdot m_{B}}$ (Eq. 2)

If we know that $\rho_{A} = 850\,\frac{kg}{m^{3}}$ , $\rho_{B} = 1060\,\frac{kg}{m^{3}}$ and $m_{A} = m_{B} = 0.075\,kg$ , then the density of the mixture:

$\rho_{M} = \frac{\left(850\,\frac{kg}{m^{3}} \right)\cdot \left(1060\,\frac{kg}{m^{3}} \right)\cdot (0.075\,kg+0.075\,kg)}{\left(1060\,\frac{kg}{m^{3}} \right)\cdot \left(0.075\,kg\right)+\left(850\,\frac{kg}{m^{3}} \right)\cdot \left(0.075\,kg\right)}$

$\rho_{M} = 943.455\,\frac{kg}{m^{3}}$

And the specific gravity is the ratio of the density of the mixture to the density of water, that is:

$SG = \frac{\rho_{M}}{\rho_{w}}$ (Eq. 3)

If we get that $\rho_{M} = 943.455\,\frac{kg}{m^{3}}$ and $\rho_{w} = 1000\,\frac{kg}{m^{3}}$ , then the specific gravity of the mixture is:

$SG = \frac{943.455\,\frac{kg}{m^{3}} }{1000\,\frac{kg}{m^{3}} }$

SG = 0.943

The specific gravity of the mixture is 0.943.

Answer from: Quest

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Answer from: Quest

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Liquid A has a mass density of 850.0 kg/m3 and Liquid B has a mass density of 1060.0 kg/m3. Seventy-...

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