/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 17 Aqueous solutions of the amino a... [FREE SOLUTION] | 91Ó°ÊÓ

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Aqueous solutions of the amino acid \(\mathrm{L}\) -isoleucine (Ile) are prepared by putting 100.0 grams of pure water into each of six flasks and adding different precisely weighed quantities of Ile to each flask. The densities of the solutions at \(50.0 \pm 0.05^{\circ} \mathrm{C}\) are then measured with a precision densitometer, with the following results: $$\begin{array}{|l|l|l|l|l|l|l|}\hline r\left(\mathrm{g} \mathrm{Ile} / 100 \mathrm{g} \mathrm{H}_{2} \mathrm{O}\right) & 0.0000 & 0.8821 & 1.7683 & 2.6412 & 3.4093 & 4.2064 \\\\\hline \rho\left(\mathrm{g} \text { solution } / \mathrm{cm}^{3}\right) & 0.98803 & 0.98984 & 0.99148 & 0.99297 & 0.99439 & 0.99580 \\\\\hline\end{array}$$ (a) Plot a calibration curve showing the mass ratio, \(r,\) as a function of solution density, \(\rho,\) and fit a straight line to the data to obtain an equation of the form \(r=a \rho+b\) (b) The volumetric flow rate of an aqueous Ile solution at a temperature of \(50^{\circ} \mathrm{C}\) is \(150 \mathrm{L} / \mathrm{h}\). The density of a sample of the stream is measured at \(50^{\circ} \mathrm{C}\) and found to be \(0.9940 \mathrm{g} / \mathrm{cm}^{3} .\) Use the calibration equation to estimate the mass flow rate of Ile in the stream (kg IIe/h). (c) It has just been discovered that the thermocouple used to measure the stream temperature was poorly calibrated and the temperature was actually \(47^{\circ} \mathrm{C}\). Would the Ile mass flow rate calculated in Part \((\mathrm{b})\) be too high or too low? State any assumption you make and briefly explain your reasoning.

Short Answer

Expert verified
Firstly, a calibration curve is plotted using the provided data and a calibration equation of the form \( r = a\rho + b \) is obtained. Then, using the volumetric flow rate, the density, and the calibration equation, the mass flow rate of Ile is calculated. Lastly, the effect of a temperature change from 50°C to 47°C on the Ile mass flow rate is considered, and it is deduced that the actual mass flow rate would most likely be lower than the calculated one due to the temperature being lower than expected.

Step by step solution

01

Plotting the data

Plot the data given by setting the mass ratio, \( r \), on the x-axis and the solution density, \( \rho \), on the y-axis. Each pair \( (r, \rho) \) represents a point on the graph.
02

Obtain the calibration equation

Perform a linear regression on the plotted points to obtain the equation of the line. This is the calibration equation of the form \( r = a\rho + b \), where \( a \) is the slope of the line and \( b \) is the y-intercept.
03

Volumetric flow rate to mass flow rate

Using the volumetric flow rate of the Ile solution and the obtained calibration equation, calculate the mass flow rate of Ile. The density of the sample and the volumetric rate are needed for this. Use the equation \( \rho = \frac{mass}{volume} \) to find the mass flow rate
04

Effect of temperature changes

Examine the effect of the change in temperature from 50°C to 47°C on the Ile mass flow rate. Considering that the density of a substance generally decreases with an increase in temperature, the actual mass flow rate could be lower than that calculated in step 3 if the temperature is lower than estimated.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Calibration Curve
When preparing solutions in a chemical lab, it's crucial to relate the physical properties of the solution with the concentration of solute present. A calibration curve is a tool that enables us to make these connections. Think of it as a graph where we plot known concentrations of our solute (in this case, the amino acid Isoleucine) against a property that changes with concentration, like density. We create this plot using precise measurements, which is why the densitometer's role in our exercise is indispensable.

With the carefully measured pairs of isoleucine concentration and solution density, the line drawn on the graph (the calibration curve) now serves as a reference. By finding any point along this line, we can predict the concentration of isoleucine in a new solution just by knowing its density. This curve is central to chemical analysis and quality control processes because it ensures the accurate determination of concentrations within various solutions.

Once the curve is established, the line's equation, in our exercise designated as \( r = a\rho + b \), where \( r \) is the mass ratio and \( a \) and \( b \) are constants derived from the graph, becomes a formula that translates density into concentration. This is crucial for maintaining consistent chemical processes, such as verifying the purity of a pharmaceutical ingredient or monitoring pollutant levels in water.
Linear Regression
Linear regression is a powerful statistical tool used to find relationships between variables. In the context of our chemical process exercise, it's employed to produce the best-fit line on a calibration curve. This line is a model that represents the relationship between solution density and the mass ratio of Ile in water.

The goal of linear regression is to find the most accurate values for the slope \( a \) and the intercept \( b \) such that the equation \( r = a\rho + b \) minimizes the difference between the observed densities and those predicted by our model. This method is widely used in science and engineering because it helps convert raw data into meaningful information. For example, if a company needs to monitor the concentration of a chemical in their product, linear regression can assist in predicting the outcomes based on a measurable property, enhancing the predictability and control over the manufacturing process.

In the linear regression process, the precision of the data points (in our case, the exact measurements of density at different concentrations) is crucial to derive a reliable and useful calibration equation. This equation becomes the basis for further calculations and quality assessments in the chemical process.
Mass Flow Rate
Mass flow rate is a term that describes how much mass passes through a given point in a system per unit time. It's a crucial concept in chemical engineering and manufacturing for controlling and optimizing processes. In the context of our exercise, we're interested in the amount of Isoleucine flowing through a system per hour. To find this, we look at the volumetric flow rate, which tells us the volume of Ile solution passing through per hour, and the solution’s density.

However, the real magic happens when we use our calibration equation derived from the calibration curve and linear regression. This equation transforms our knowledge of the solution's density into an estimation of the Isoleucine concentration. Combining this with the volumetric flow rate and the known density, we finally obtain the mass flow rate of Isoleucine. It's worth noting that mass flow rate is influenced by numerous factors such as pressure, temperature, and the physical properties of the fluid. All these need to be accounted for a correct estimation, as seen in our exercise when the need arose to adjust for an incorrectly measured temperature.

The precision of understanding mass flow rate has profound implications, from ensuring the proper dosing in pharmaceuticals to maintaining the safety and efficiency of food processing. It exemplifies how intricate and necessary it is to measure and control each step of a chemical process with as much accuracy as possible.

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Most popular questions from this chapter

As will be discussed in detail in Chapter \(5,\) the ideal-gas equation of state relates absolute pressure, \(P(\mathrm{atm}) ;\) gas volume, \(V(\text { liters }) ;\) number of moles of gas, \(n(\mathrm{mol}) ;\) and absolute temperature, \(T(\mathrm{K}):\) $$P V=0.08206 n T$$ (a) Convert the equation to one relating \(P(\mathrm{psig}), V\left(\mathrm{ft}^{3}\right), n(\mathrm{lb}-\mathrm{mole}),\) and \(T\left(^{\circ} \mathbf{F}\right)\). (b) \(\mathrm{A} 30.0\) mole \(\%\) CO and 70.0 mole \(\% \mathrm{N}_{2}\) gas mixture is stored in a cylinder with a volume of \(3.5 \mathrm{ft}^{3}\) at a temperature of \(85^{\circ} \mathrm{F}\). The reading on a Bourdon gauge attached to the cylinder is 500 psi. Calculate the total amount of gas (lb- mole) and the mass of \(\mathrm{CO}\left(\mathrm{Ib}_{\mathrm{m}}\right)\) in the tank. (c) Approximately to what temperature \(\left(^{\circ} \mathrm{F}\right)\) would the cylinder have to be heated to increase the gas pressure to 3000 psig, the rated safety limit of the cylinder? (The estimate would only be approximate because the ideal gas equation of state would not be accurate at pressures this high.)

A useful measure of an individual's physical condition is the fraction of his or her body that consists of fat. This problem describes a simple technique for estimating this fraction by weighing the individual twice, once in air and once submerged in water. (a) A man has body mass \(m_{b}=122.5 \mathrm{kg} .\) If he stands on a scale calibrated to read in newtons, what would the reading be? If he then stands on a scale while he is totally submerged in water at \(30^{\circ} \mathrm{C}\) (specific gravity \(=0.996\) ) and the scale reads \(44.0 \mathrm{N},\) what is the volume of his body (liters)? (Hint: Recall from Archimedes' principle that the weight of a submerged object equals the weight in air minus the buoyant force on the object, which in turn equals the weight of water displaced by the object. Neglect the buoyant force of air.) What is his body density, \(\rho_{\mathrm{b}}(\mathrm{kg} / \mathrm{L}) ?\) (b) Suppose the body is divided into fat and nonfat components, and that \(x_{f}\) (kilograms of fat/kilogram of total body mass) is the fraction of the total body mass that is fat: \(x_{\mathrm{f}}=\frac{m_{\mathrm{f}}}{m_{\mathrm{b}}}\) Prove that \(x_{\mathrm{f}}=\frac{\frac{1}{\rho_{\mathrm{b}}}-\frac{1}{\rho_{\mathrm{nf}}}}{\frac{1}{\rho_{\mathrm{f}}}-\frac{1}{\rho_{\mathrm{nf}}}}\) where \(\rho_{\mathrm{b}}, \rho_{\mathrm{f}},\) and \(\rho_{\mathrm{nf}}\) are the average densities of the whole body, the fat component, and the nonfat component, respectively. [Suggestion: Start by labeling the masses ( \(m_{\mathrm{f}}\) and \(m_{\mathrm{b}}\) ) and volumes \(\left(V_{\mathrm{f}} \text { and } V_{\mathrm{b}}\right)\) of the fat component of the body and the whole body, and then write expressions for the three densities in terms of these quantities. Then eliminate volumes algebraically and obtain an expression for \(\left.m_{f} / m_{b} \text { in terms of the densities. }\right]\) (c) If the average specific gravity of body fat is 0.9 and that of nonfat tissue is \(1.1,\) what fraction of the man's body in Part (a) consists of fat? (d) The body volume calculated in Part (a) includes volumes occupied by gas in the digestive tract, sinuses, and lungs. The sum of the first two volumes is roughly \(100 \mathrm{mL}\) and the volume of the lungs is roughly 1.2 liters. The mass of the gas is negligible. Use this information to improve your estimate of \(x_{\mathrm{f}}\).

A thermostat control with dial markings from 0 to 100 is used to regulate the temperature of an oil bath. A calibration plot on logarithmic coordinates of the temperature, \(T(\text { ( } \mathrm{F} \text { ), versus the dial setting, } R\), is a straight line that passes through the points \(\left(R_{1}=20.0, T_{1}=110.0^{\circ} \mathrm{F}\right)\) and \(\left(R_{2}=40.0, T_{2}=250.0^{\circ} \mathrm{F}.\right)\) (a) Derive an equation for \(T\left(^{\circ} \mathrm{F}\right)\) in terms of \(R\). (b) Estimate the thermostat setting needed to obtain a temperature of \(320^{\circ} \mathrm{F}\). (c) Suppose you set the thermostat to the value of \(R\) calculated in Part (b), and the reading of athermocouple mounted in the bath equilibrates at \(295^{\circ} \mathrm{F}\) instead of \(320^{\circ} \mathrm{F}\). Suggest several possible explanations.

A mixture is 10.0 mole \(\%\) methyl alcohol, 75.0 mole \(\%\) methyl acetate \(\left(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}_{2}\right),\) and 15.0 mole \(\%\) acetic acid. Calculate the mass fractions of each compound. What is the average molecular weight of the mixture? What would be the mass (kg) of a sample containing 25.0 kmol of methyl acetate?

Convert the temperatures in Parts (a) and (b) and temperature intervals in Parts (c) and (d): (a) \(T=85^{\circ} \mathrm{F}\) to \(^{\circ} \mathrm{R},^{\circ} \mathrm{C}, \mathrm{K}\) (b) \(T=-10^{\circ} \mathrm{C}\) to \(\mathrm{K},^{\circ} \mathrm{F},^{\circ} \mathrm{R}\) (c) \(\Delta T=85^{\circ} \mathrm{C}\) to \(\mathrm{K},^{\circ} \mathrm{F},^{\circ} \mathrm{R}\) (d) \(\Delta T=150^{\circ} \mathrm{R}\) to \(^{\circ} \mathrm{F},^{\circ} \mathrm{C}, \mathrm{K}\)

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