How many grams of ethanol (density 0.80 g/cm³)should be added to 5.0 g of chloroform (density 1.5 g/cm³)if the resulting mixture is to have a density of 1.2 g/cm³? Assume that the fluids do not change their volumes when they are mixed. A)2.0 g B)2.4 g C)1.8 g D)4.4 g E)1.6 g

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The correct answer is A.The density of the mixture is given by:$$ho = \frac{m_{ethanol} + m_{chloroform}}{V_{ethanol} + V_{chloroform}}$$where $ho$ is the density of the mixture, $m_{ethanol}$ is the mass of ethanol, $m_{chloroform}$ is the mass of chloroform, $V_{ethanol}$ is the volume of ethanol, and $V_{chloroform}$ is the volume of chloroform.We can assume that the volumes of the fluids do not change when they are mixed, so:$$V_{ethanol} + V_{chloroform} = V_{mixture}$$where $V_{mixture}$ is the volume of the mixture.We can also write:$$V_{ethanol} = \frac{m_{ethanol}}{ho_{ethanol}}$$and$$V_{chloroform} = \frac{m_{chloroform}}{ho_{chloroform}}$$where $ho_{ethanol}$ is the density of ethanol and $ho_{chloroform}$ is the density of chloroform.Substituting these expressions into the equation for the density of the mixture, we get:$$ho = \frac{m_{ethanol} + m_{chloroform}}{\frac{m_{ethanol}}{ho_{ethanol}} + \frac{m_{chloroform}}{ho_{chloroform}}}$$Solving for $m_{ethanol}$, we get:$$m_{ethanol} = \frac{ho_{mixture} - ho_{chloroform}}{ho_{ethanol} - ho_{mixture}} 	imes m_{chloroform}$$Substituting the given values, we get:$$m_{ethanol} = \frac{1.2 g/cm^3 - 1.5 g/cm^3}{0.80 g/cm^3 - 1.2 g/cm^3} 	imes 5.0 g = 2.0 g$$Therefore, the correct answer is A.

How Many Grams of Ethanol (Density 0