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)
    Solution 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)