Chapter 29 Multiple choice questions

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. For strong electrolytes, it has been shown that the following empirical law relates the molar conductivity (L) to the concentration (c):
$\Lambda ={\Lambda }^{\infty }-b\sqrt{c}$, where ${\Lambda }^{\infty }$ is the molar conductivity at infinite dilution and b is a constant specific to the electrolyte.
What variables should be plotted on a graph to linearize the equation and what information can be extracted from the graph?

y = $\Lambda$, x = $c$, gradient = b, and intercept = ${\Lambda }^{\infty }$. correct incorrect

y = $\Lambda$, x = $\sqrt{c}$, gradient = ${\Lambda }^{\infty }$, and intercept = -b. correct incorrect

y = $\Lambda$, x = ${\Lambda }^{\infty }$, gradient = 1, and intercept = -b$\sqrt{c}$. correct incorrect

y = $\Lambda$, x = $\sqrt{c}$, gradient = -b, and intercept = ${\Lambda }^{\infty }$. correct incorrect

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. Blagden's law relates the lowering in the freezing point temperature of a solvent to the amount of dissolved solute:
$\Delta T=K_{\text{cryoscopic}}\prime \frac{mas{s}_{\text{solute}}}{M_R}\prime \frac{1000}{mas{s}_{\text{solvent}}}$.
If we are given a set of data for the decrease in the melting point, when a specific mass of the same solute is added to different masses of solvent, how should we linearize the equation above?

y = $\Delta T$, x = $\frac{1}{mas{s}_{solvent}}$, gradient = $\left(\frac{1000K_{\text{cryoscopic}}\times mas{s}_{\text{solute}}}{M_R}\right)$, and the intercept = 0. correct incorrect

y = $\frac{1}{mas{s}_{\text{solvent}}}$, x = $\Delta T$, gradient = $\left(\frac{1000K_{\text{cryoscopic}}\times mas{s}_{\text{solute}}}{M_R}\right)$, and the intercept = 0. correct incorrect

y = $\Delta T$, x = $\frac{1}{mas{s}_{\text{solvent}}}$, gradient = $\left(\frac{1000mas{s}_{\text{solute}}}{M_R}\right)$, and the intercept = $K_{\text{cryoscopic}}$. correct incorrect

y = $\Delta T$, x = $\left(\frac{mas{s}_{\text{solute}}}{M_R}\right)$, gradient = $K_{\text{cryoscopic}}$, and the intercept = $\frac{1000}{mas{s}_{\text{solvent}}}$. correct incorrect

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. We are given a set of data (x, y), which is thought to fit the equation: $y=a{x}^b$. Select the correct form from the list below which set of axes enables us to find the values of a and b.

Plot y on the y-axis, x on the x-axis. The data will lie on a straight line, the gradient of which will be a and the intercept will be b. correct incorrect

Plot ln y on the y-axis, ln x on the x-axis. The data will lie on a straight line, the gradient of which will be b and the intercept will be ln a. correct incorrect

Plot ln y on the y-axis, x on the x-axis. The data will lie on a straight line, the gradient of which will be ln b and the intercept will be ln a. correct incorrect

Plot y on the y-axis, ln x on the x-axis. The data will lie on a straight line, the gradient of which will be b and the intercept will be ln a. correct incorrect

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. We are given a set of data (x, y), which is thought to fit the equation: $y=a{b}^x$. Select the correct form from the list below which enables us to find the values of a and b.

Plot ln y on the y-axis, ln x on the x-axis. The data will lie on a straight line, the gradient of which will be b and the intercept will be ln a. correct incorrect

Plot y on the y-axis, x on the x-axis. The data will lie on a straight line, the gradient of which will be a and the intercept will be b. correct incorrect

Plot y on the y-axis, ln x on the x-axis. The data will lie on a straight line, the gradient of which will be ln b and the intercept will be ln a. correct incorrect

Plot ln y on the y-axis and x on the x-axis. The data will lie on a straight line, the gradient of which will be ln b and the intercept will be ln a. correct incorrect

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. We are given a set of data (x, y), which are thought to fit the equation: $y=a{x}^b$.

x	1.2	2.5	3.6	4.2	6.3
y	3.97	35.94	107.31	170.40	575.11

Find the values of a and b.

a = 0.832 and b = 3.0 correct incorrect

a = 0.832 and b = 20.1 correct incorrect

a = 20.1 and b = 0.832 correct incorrect

a = 2.3 and b = 3.0 correct incorrect

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. We are given a set of data (x, y), which are thought to fit the equation: $y=a{b}^x$.

x	1.2	2.5	3.6	4.2	6.3
y	8.4	44.3	181.1	390.6	5754.1

Find the values of a and b.

a = 1.8 and b = 1.3 correct incorrect

a = 1.8 and b = 3.6 correct incorrect

a = 0.6 and b = 1.3 correct incorrect

a = 0.6 and b = 3.6 correct incorrect

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. The data in the table below were collected for a reaction known to follow a second-order kinetic rate law.

t /s	30	60	120	180	240
[A]t / mol dm−3	0.0446	0.0226	0.0114	0.0076	0.0057

Plot a suitable graph in order to determine the rate constant for the reaction, k.

k = 0.45 dm³ mol^-1 s^-1 correct incorrect

k = 0.04 dm³ mol^-1 s^-1 correct incorrect

k = 0.73 dm³ mol^-1 s^-1 correct incorrect

k = 0.00016 dm³ mol^-1 s^-1 correct incorrect

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. We are given the following set of data for a half-cell reaction at 298 K: Xⁿ⁺ + ne^- ? X.

[Xn+]/mol dm−3	0.5	0.2	0.1	0.05	0.01
$E_{{\mathrm{X}}^{n+},\mathrm{X}}/\mathrm{V}$	0.556	0.548	0.542	0.536	0.523

Find the number of electrons n in the half-cell reaction.
(R = 8.314 J K^-1 mol^-1, F = 96485 C mol^-1)

n = 3 correct incorrect

n = 2 correct incorrect

n = 1 correct incorrect

n = 1.5 correct incorrect

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. The mass of a compound A that is adsorbed onto a surface is related to its concentration at equilibrium by the Freundlich adsorption isotherm: $m=k{c}^{\frac{1}{n}}$.
We have been given the following set of data:

c/ 10–4 mol dm–3	2.4	3.9	14.3	28.2	36.5
m/ 10–3 mol	6.03	6.77	9.23	10.8	11.5

Plot a suitable graph to determine the values for k and n.

k = -3.136 and n = 4.22 correct incorrect

k = -3.136 and n = 0.24 correct incorrect

k = 0.043 and n = 0.24 correct incorrect

k = 0.043 and n = 4.22 correct incorrect

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. We are given the following information about a radioactive isotope which is decomposing:

t/min	5	10	15	20	25
nt /mol	1.80	1.62	1.45	1.31	1.17

where n_t is the amount of radioactive isotope remaining at time t.
Using the equation to predict half-lives given in Chapter 29, plot a suitable graph to determine n₀ and the half-life of the material, t_½.

n₀ = 2 mol and t_½= 0.02 mins correct incorrect

n₀ = 2 mol and t_½= 32 mins correct incorrect

n₀ = 2 mol and t_½= 32 min correct incorrect

n₀ = 2 mol and t_½= 47 mins correct incorrect