Question

11A.2 What is (a) the wavenumber, (b) the wavelength of the radiation used by an FM radio transmitter broadcasting at 88.0 MHz? 11A.3 An aqueous solution of a triphosphate derivative of molar mass 502 g mol-' was prepared by dissolving 17.2 mg in enough water to make 500 cm of solution and a sample was transferred to a cell of length 1.00 cm. The absorbance was measured as 1.011. Calculate (a) the molar absorption coefficient; (b) the transmittance, expressed as a percentage, for a solution of twice the concentration.

Answer 1

T1-A. 2) Given frequency = v = 88.0 MHZ TMHZ =88.ox = 1 fHZ 10GHZ v The expression for wave No (v) is shown below o = 1 / at wavelength We know that =- c = speed of light = 3x10 m/s ut frequency = 88.0x10542 88.0810 Hz 3x108 m/s - 29. 23 x 102 - = 8.933 X 1o'm' 0.2933m Answer i 0.2933m) - 1 a = 3.41m Im = loom = 341CM

(14.3 Amount of solute = m = 17.2mg = 0.0172g 'g = lodong o - Molar mass = 502g/mole No of moles of solute = 0.0172g - 3.428105 male - 50zglmole" Molas masso - * volume of solution = v= 500cm3 V = 5oomL = 0.5L - 1m = IML 1L = 1000mL Concentration of solution = c = D - 3-42x 10 mot o.SL O c = 6.84 x 10 mole/L c = 6.84 x 10 mol/dm3 Given Absorbance = A = 1.011 IL = 1 dm3 we know - H=ed E = Molar absorption coefficient l = length of cell =1.00cm o n A = ed 1-oll = EX 6-84 X10 Smoll dm3 x de 1-00) & = 0.1478 x 105 L moicnt = 1.478 x 104 L. moi on Anone

i (b) The relation Absorbance between and The delan Transmittance A = log / T = Transmittance 11.011 = log (th - Taking antilog both sides we get I = 1011.011 = 10.25 L T - 0.0975 1% T = loox T - 9.75% I Answer