A mixture contains 23 g of ethanol (MW = 46) and 36 g of water. Some acetic acid (MW = 60) is added to the mixture, and its mole fraction becomes 0.2. How much acetic acid was adde... A mixture contains 23 g of ethanol (MW = 46) and 36 g of water. Some acetic acid (MW = 60) is added to the mixture, and its mole fraction becomes 0.2. How much acetic acid was added? Also, derive the equation of hydrostatic pressure.
Understand the Problem
The image presents two chemistry questions. The first question (Q2/A) involves calculating the mass of acetic acid added to a mixture of ethanol and water, given the final mole fraction of acetic acid. The second question (Q2/B) requires deriving the equation for hydrostatic pressure.
Answer
Q2/A: $37.5 \text{ g}$ Q2/B: $P = P_0 + \rho g h$
Answer for screen readers
Q2/A: The mass of acetic acid added is $37.5 \text{ g}$. Q2/B: The equation for hydrostatic pressure is $P = P_0 + \rho g h$ or $P = \rho g h$.
Steps to Solve
Question A
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Calculate moles of ethanol Moles of ethanol = mass / molecular weight = $23 \text{ g} / 46 \text{ g/mol} = 0.5 \text{ mol}$
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Calculate moles of water Moles of water = mass / molecular weight = $36 \text{ g} / 18 \text{ g/mol} = 2 \text{ mol}$
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Define variables and set up the mole fraction equation Let $x$ be the moles of acetic acid added. The mole fraction of acetic acid is given by: $$ \frac{x}{x + 0.5 + 2} = 0.2 $$
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Solve for x (moles of acetic acid) $x = 0.2(x + 2.5)$ $x = 0.2x + 0.5$ $0.8x = 0.5$ $x = 0.5 / 0.8 = 0.625 \text{ mol}$
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Calculate mass of acetic acid Mass of acetic acid = moles * molecular weight = $0.625 \text{ mol} * 60 \text{ g/mol} = 37.5 \text{ g}$
Question B
- Consider a small area within a fluid
Consider a small horizontal area $dA$ at a depth $h$ below the surface of a fluid, density $\rho$
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Calculate the weight of the fluid column above the area The volume of the fluid column above the area is $dV = dA \cdot dh$, where $dh$ is the height of the fluid column. The mass of the fluid column is $dm = \rho \cdot dV = \rho \cdot dA \cdot dh$. The weight of the fluid column is $dW = dm \cdot g = \rho \cdot g \cdot dA \cdot dh$, where $g$ is the acceleration due to gravity.
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Calculate the pressure exerted by this column of fluid. Pressure is force per unit area, $P = \frac{F}{A}$. Considering that the force is the weight $dW$, $$ dP = \frac{dW}{dA} = \frac{\rho g dA dh}{dA} = \rho g dh $$
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Integrate to find total pressure Integrate the pressure differential from the surface ($h = 0$) to depth h: $$ \int_{P_0}^{P} dP = \int_{0}^{h} \rho g dh $$ Since $\rho$ and $g$ are constant: $$ P - P_0 = \rho g h $$ where $P_0$ is the atmospheric pressure at the surface.
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Final equation for hydrostatic pressure $P = P_0 + \rho g h$ If we are only interested in the pressure due to the fluid: $P = \rho g h$
Q2/A: The mass of acetic acid added is $37.5 \text{ g}$. Q2/B: The equation for hydrostatic pressure is $P = P_0 + \rho g h$ or $P = \rho g h$.
More Information
The hydrostatic pressure equation shows that pressure increases linearly with depth in a fluid. This is why dams are thicker at the bottom than at the top, to withstand the greater pressure at greater depths.
Tips
For Q2/A, a common mistake is not including the moles of acetic acid in the denominator when calculating the mole fraction, leading to an incorrect equation and result. Another mistake is to forget multiplying the moles of acetic acid by its molecular weight to obtain mass asked by the questions. For Q2/B, a common mistake is forgetting to include atmospheric pressure $P_0$ in the final equation. Also, not understanding the integral relation between pressure differential and height.
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