LAB_4 last year

# LAB_4 last year - Instrumental Measurement Laboratory...

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Instrumental Measurement Laboratory CHEE291 Department of Chemical Engineering McGill University Chemical Analysis TA for this lab: Ms. Lindsay Lessard Code number of the unknown sample is (2A)

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CALCULATIONS A. Maximum soluble [NaNO 3 ] in mol/L solution: Since 1 liter/x liter = 1000 cm 3 / 1 cm 3 Therefore x = 1 L.cm 3 / 1000 cm 3 = 0.001L Solubility of NaNO 3 = 92.1g/ 100cm 3 H 2 O = 92.1g /0.1 L H 2 O = 921 g/ L Molecular weight of NaNO 3 = 84.99g/mol [NaNO 3 ] = (921 g/L)*(1 mol/ 84.99g) = 921 mol/84.99 L = 10.84mol/L B. To find the maximum possible (soluble) concentration of CuSO 4 in mol/L First, need to find the solubility of CuSO 4 at 25˚C Interpolation: (Solubility of CuSO 4 at 25˚C) = [31.6g/0.1L+ (25˚C -0˚C/100˚C -0˚C)*(203.3 g/0.1L -31.6g/0.1L)] = 74.53g/0.1L = 745.3g/L Concentration of CuSO 4 in 1mol/1L (745.3g/L)[(1mol CuSo 4 .5H 2 0)/(1mol of CusO 4 )](1mol/249.68g) = 2.99mol/L C. Concentration: (0.5/2) mol/L = 0.25 mol/L Volume: 500 ml = 0.5 L # moles of CuSO 4 / total Volume = [CuSO 4 ] # moles of CuSO 4 = (0.25)*(0.5) = 0.125 mol Molecular weight of CuSO 4 .5 H 2 O = 249.68g/mol # mol = m/M Therefore, m = (#mol)*(M) = (0.125mol)*(249.68g/mol) = 31.21g D . The initial volumes for the solution of CuSO 4 C 1 V 1 = C 2 V 2 V 1 : unknown Therefore V 1 =C 2 V 2 / C 1 V 2 = 100 ml =0.1 L C 1 = (0.5/2) molCuSO 4 /1L = 0.25 molCuSO 4 /1L C 2 = (0.1/2) molCuSO 4 /1L =0.05 molCuSO 4 /1L V 1 = 0.02L (0.2/2) molCuSO 4 /1L = 0.1molCuSO 4 /1L V 1 = 0.04L (0.3/2) molCuSO 4 /1L = 0.15 molCuSO 4 /1L V 1 = 0.06L (0.4/2) molCuSO 4 /1L = 0.2 molCuSO 4 /1L V 1 = 0.08L Example with the first C 2: V 1 =C 2 V 2 / C 1 = (0.05)*(0.1)/(0.25) = 0.02L 2
RESULTS Some Calculation within the Experimental procedures In the spectrophotometer, the maximum absorbance occurs at a wavelength of 800 nm. 6) Absorbance and [CuSO

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LAB_4 last year - Instrumental Measurement Laboratory...

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