# PHYS*1080 Sample Quiz 10

1. A spherical molecule of density $1.2 \times 10^3 \; \text{kg/m}^3$ and a molar mass of $4.0 \times 10^5 \; \text{g/mol}$ is in an ultracentridfuge tube (filled with water at $20^\circ \text C$) at a distance of $15 \; \text{cm}$ from the rotation axis. The rotor is spinning at $6000\; \text{Hz}.$ How long does it take the molecule to sediment $2.0\; \mu m$ ? (5 marks)

Solution Marks
$v_t = \frac{8(\pi ^2 f^2R)r^2(\rho_s-\rho_\ell)}{9 \eta}$ (1.0) formula
$d = vt$ (0.5) $d$
$t = \frac{d}{v_t} = \frac{d9 \eta}{8(\pi^2f^2R)r^2(\rho_s- \rho_\ell)}$ (0.5) $t$
$N_A \frac{4}{3}\pi r^3\rho_s = M$ (1.0) $M$
\begin {align} r^3 & = \frac{4 \times 10^2 \times 3}{6.02 \times 10^{23}\times 4 \times \pi \times 1.2 \times 10^3} \\ & = 1.32 \times 10^{-25} \\ r^2 & = 2.59 \times 10^{-17} \text m^2 \end {align} (1.0) $r^2$
\begin {align} t & = \frac{2 \times 10^{-6} \times 9 \times 1 \times 10^{-3}}{8 \times \pi^2 \times 36 \times 10^6\times 0.15 \times 0.2 \times 10^3 \times 2.59 \times 10^{-17}} \\ & = 8.2 \times 10^{-3} \text s \end {align} (1.0) answer

2. In the simulated diffusion experiment, if you had concentration measurements ranging form $0.5$ to $0.005$ over a period of time what kind of graph paper would you require to plot all the experimental points to yiled a straight line? (2 marks)

Solution Marks
3 cycle semi-log paper is required.
$0.005$ (part of first cycle) $\rightarrow 0.01 \rightarrow 0.1$(second cycle) $\rightarrow$
(1.0) for 3 cycle (0.5 for 2 cycle )
$0.5$ (part of the third cycle) (1.0) for semi-log

3. The diffusion coefficient of a molecule in water at $10^\circ \text C$ is $4.3 \times 10^{-10}\; \text m^2/\text s.$ At this temperature the viscosity of water is $1.307 \times 10^{-3}\; Pa\; \text s$. What will be the diffusion coefficient at $50^\circ \text C$ where the viscosity of water is $0.547 \times 10^{-3}\; Pa\; \text s$ ? (3 marks)

$D = \frac{kT}{6 \pi \eta r} \quad i.e., \; D\; \alpha \frac{T}{\eta}$ (1.0)
$D_{50} = D_{10} \times \frac{323}{283} \times \frac{1.307 \times 10^{-3}}{0.547 \times 10^{-3}}$ (1.0) (0.5, for each ratio)
$= 1.2 \times 10^{-9}\; \text{m}^2/ \text s$ (1.0)