Chemistry Labs

International Chemistry Olympiad · 2023

Problems

  1. Problem 1Molecular imaging is a powerful tool in medical diagnostics. The nuclear isomer 99mTc^{99m}\ce{Tc} (m = metastable) is an excellent γ-emitter (t1/2=6.015t_{1/2} = 6.015 h) obtained by β− decay of a mother nuclide in a technetium generator as [99mTcOX4]−[^{99m}\ce{TcO4}]^-. (a) Identify the mother nuclide A and the emitted particle B in A→X99mX2299mTc+B\ce{A -> ^{99m}Tc + B}. (b) An eluate from a 99mTc^{99m}\ce{Tc} generator has an activity of 12.5 GBq (1 GBq = 10910^{9} decays per second). Calculate how many moles of 99mTc^{99m}\ce{Tc} are present. (c) For standard imaging about 200 MBq are administered to a patient. Assuming no activity is lost through excretion, calculate how many hours the patient has to wait until the injected activity decreases to under 1% of the starting activity.Solutions: 1
  2. Problem 5To remove sulfur from fuels, hydrogen-assisted hydrodesulfurization is used at refineries, typically over MoSX2\ce{MoS2} supported on SiOX2\ce{SiO2}. Isotope exchange at the gas–solid interface exchanges only the surface atoms. An experiment studies the exchange between an MoSX2/SiOX2\ce{MoS2/SiO2} catalyst (mcat=1.2350m_{\text{cat}} = 1.2350 g, Mo mass fraction wMo=4.280%w_{\ce{Mo}} = 4.280\%, initially containing only 32^{32}S) and gaseous HX2X34X2234S\ce{H2^{34}S} in a flow reactor (p=1.00p = 1.00 bar, flow v=20.0v = 20.0 mL min−1^{-1}, T=23.0T = 23.0 °C, φ(HX2X34X2234S)=1.00%\varphi(\ce{H2^{34}S}) = 1.00\%, isotopic purity α=99.95\alpha = 99.95 mol%). After t=10.0t = 10.0 min, the fraction of 34^{34}S among sulfur atoms in the collected gas was γ=87.3\gamma = 87.3 mol%. (a) Calculate the amount of exchanged (surface) sulfur atoms n(S)surfacen(\mathrm{S})_{\text{surface}} in mol. (b) Assuming uniform spherical MoSX2\ce{MoS2} particles of density ρ=5.06\rho = 5.06 g cm−3^{-3}, surface areas per atom AS=3.00×10−19A_S = 3.00\times 10^{-19} m² (S) and AMo=5.00×10−19A_{\ce{Mo}} = 5.00\times 10^{-19} m² (Mo), and that only half of each MoSX2\ce{MoS2} unit is exposed at the surface, calculate the particle radius RR in nm (M(MoSX2)=160.07M(\ce{MoS2}) = 160.07, M(Mo)=95.95M(\ce{Mo}) = 95.95 g mol−1^{-1}).Solutions: 1