Exper i_Density of water

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Experiment (i): Statistical Analysis of Data: Standard Deviation, Confidence Limit, and Outliers (1) Abstract The density of distilled water was calculated at standard temperature and pressure. Mass and volume of water samples were found using a analytical balance and volumetric cylinder. Statistical analysis was performed to calculate mean mass, volume, and hence density. Density calculated from these mean values was ρ = ( 0.98 ± 0.27 ) g / ml , where the confidence limit is a

propagated error of the density. Density was also calculated using the LINEST function in Excel, which gave ρ = ( 0.97 ± 0.023 ) g / ml . (2) Introduction The objective of the experiment is to determine the density of distilled water. The experiment is conducted by measuring mass with an analytical balance and volume using a graduated cylinder. The density can then be calculated from the ratio of the changes in mass and volume as drops of water are added. Statistical analysis will be performed. (3) Theory Density is a measurement of mass per unit volume (1). Density is a physical property. (4) Experimental Distilled water was dropped into a 10 mL graduated cylinder until a volume of 1.00 mL was reached. The graduated cylinder was placed on the analytical balance and the balance was then tarred. The following procedure was repeated a total of 9 times: the graduated cylinder was removed from the balance and 10 drops of water were added using an eyedropper; the volume as read from the graduated cylinder and the mass as read from the balance were recorded. Only after a total of 9 trials of adding 10 drops was the graduated cylinder emptied. (5) Results Table 1. Mass and Volume of Water Drop Trials Trial V, ml ΔV, ml m, g Δm, g 0 1.00 0 1 1.50 0.50 0.5364 0.5364 2 1.90 0.40 0.9600 0.4236 3 2.30 0.40 1.3600 0.4000 4 3.00 0.70 1.9597 0.5997 5 3.40 0.40 2.4066 0.4469 6 3.90 0.50 2.8430 0.4364 7 4.30 0.40 3.2799 0.4369 8 4.90 0.60 3.7998 0.5199 9 5.30 0.40 4.2125 0.4127 Table 2. Statistical Analysis of Changes in Mass and Volume ΔV, ml Δm, g Mean 0.48 0.4681 St dev, S x 0.11 0.0678 Error of mean, δ x 0.036 0.0226

CL 0.084 0.0522 Table 3. Calculated density from Mean Values and LINEST ϱ±CL (from mean ∆ m and ∆V ), g/ml ϱ±CL (from plot) g/ml 0.98±0.27 0.97±0.023 Figure 1. Density of Water from Slope of Graph 0.00 1.00 2.00 3.00 4.00 5.00 6.00 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 f(x) = 0.97 x − 0.91 R² = 1 Mass vs. Volume for Water samples Volume [ml] Mass [g] Table 4. LINEST Excel Data slope,m 0.968488 intercept -0.91495 std error of m 0.0082 std error of int 0.028227 r^2 0.999427 std error of y 0.036008 f 13950.91 degree of freedom 8 regression ss 18.08874 residual SS 0.010373 Q-test for outliers on data for ∆ V and ∆ m. Start by sorting data lowest to highest. Q = | suspect value − nearest value | | largest value − smallest value | Q ∆V 1 = | 0.40 − 0.50 | ml | 0.70 − 0.40 | ml Q ∆V 1 = 0.33

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Q ∆V 2 = | 0.70 − 0.60 | ml | 0.70 − 0.40 | ml Q ∆V 2 = 0.33 The critical value for N=9 measurements (of V or deltam) using a 95% CL is Qcrit=0.493. Q<Qcrit so no outliers exist for the extreme values of mass and volume changes. Q ∆m 1 = | 0.4000 − 0.4127 | g | 0.5997 − 0.4000 | g Q ∆m 1 = 0.0636 Q ∆m 2 = | 0.5997 − 0.5364 | g | 0.5997 − 0.4000 | g Q ∆m 2 = 0.3170 Mean mass and volume using ∆ V and ∆ m X = X 1 + X 2 + … + X n n ∆V = ( 0.5 + 0.4 + 0.40 + 0.70 + 0.40 + 0.50 + 0.40 + 0.60 + 0.40 ) ml 9 ∆V = 0.48 ml ∆ m = ( 0.5364 + 0.9600 + 1.3600 + 1.9597 + 2.4066 + 2.8430 + 3.2799 + 3.7998 + 4.2125 ) g 9 ∆ m = 0.4681 g Standard deviation of volume and mass differences s x = √ 1 n − 1 ∑ t = 1 n ( X t − X ) 2 s ∆V = 0.11 ml s ∆ m = 0.0678 g Standard error of the mean volume and mass, using n=9 values δX = s x √ n δV = 0.11 ml √ 9 δV = 0.036 ml

δm = 0.0678 g √ 9 δm = 0.0226 g Confidence limit using previously found s x , n=9. Using t-value for 95% confidence interval t=2.31 CL = ±t ∙ s x √ n = ±t ∙δX C L V = ± 2.31 ∙ 0.11 ml √ 9 C L V = ± 0.084 ml C L m = ± 2.31 ∙ 0.0678 g √ 9 C L m = ± 0.0522 g Density of water using mean volume and mass ρ = m / V ρ = 0.4681 g 0.48 ml ρ = 0.98 g / ml Propagation of error for calculated density using average and standard deviations of m and V. m ave = ( 0.4681 ± 0.07 ) g and V ave = ( 0.48 ± 0.11 ) ml δρ = | ρ | ∙ √ ( δm m ) 2 + ( δV V ) 2 δρ = | 0.98 g / ml | ∙ √ ( 0.07 g 0.4681 g ) 2 + ( 0.11 ml 0.48 ml ) 2 δρ = 0.27 g / ml So density calculated from the mean should be reported as ρ = ( 0.98 ± 0.27 ) g / ml Density from LINEST ρ = 0.97 g / ml Standard error of the slope S m =0.0082. The critical value for a 95% CI is found in excel to be 2.751524. CI = t critical ∙S m

CI = 2.751524 ∙ 0.0082 CI =0.023 g/ml , So ρ = ( 0.97 ± 0.023 ) g / ml using LINEST based on the graph of density. Volume of 1 drop of water. ∆ V is based on n=10 drops. V 1 = V ave 10 V 1 = 0.48 ml 10 V 1 = 0.048 ml Propagation of error for volume of 1 drop. δ V 1 = | − 1 n | δ V 10 δ V 1 = | − 1 10 | 0.036 ml δ V 1 = 0.0036 ml Number of moles in 1 drop of water. Find mass then divide by molecular weight of 18.01 g/mol (3). m 1 = m ave 10 m 1 = 0.4681 g 10 m 1 = 0.04681 g n 1 = m 1 MW n 1 = 0.04681 g 18.01 g / mol n 1 = 0.002599 mol per drop Calculate number of moles of H20 needed to have 0.5 atm vapor pressure in 1 liter. Use R=0.082057 L atm/(mol K) (2). n = PV RT

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n = ( 0.5 atm ) ∗ 1 L 0.082057 Latm mol K ∗ 273.15 K n = 0.0223 mol H 2 O needed drops = moles needed moles ∈ 1 drop,n 1 drops = 8.6 drops H 2 0 ¿ acheive 0.5 atmvapor pressure for 1 L (6) Discussion The experimental procedure consisted of performing 9 trials of adding 10 drops of distilled water to a graduated cylinder. From the experimentally determined changes in mass and volume, a mean change in mass and volume was determined. These mean values were used to calculate the density, as mass divided by volume. Propagation of the error on the density calculated using the density formula was performed. Density calculated from these mean values should be reported as ρ = ( 0.98 ± 0.27 ) g / ml , where the confidence limit is a propagated error of the density. Density was also calculated using the LINEST function in Excel, which performs linear regression using the method of least squares (1). In this case, the values of mass and volume as read experimentally were used, not the changes. So ρ = ( 0.97 ± 0.023 ) g / ml using LINEST method. The same value for density was obtained by graphing mass vs. volume, so that the slope is ρ =0.97. The literature value for density of water at 25°C is 0.997 g/ml (4). This is within 2% accuracy of the experimental results. The assumption of a 25°C room is a experimental limitation that prevents true analysis of the results, as the appropriate reference value for density may be higher or lower depending on the true room temperature. Nevertheless, this error is expected to be small as density of water ranges from 0.998 g/ml at 25°C to 0.996 g/ml at 30°C. (7) References 1. “Laboratory Manual” Analytical/ Physical Chemistry Lab (CHEM 339). Department of Chemistry and Environmental Science. New Jersey Institute of Technology. Newark, NJ, 07102. Accessed 24 May 2023. 2. https://chem.libretexts.org/Bookshelves/ Physical_and_Theoretical_Chemistry_Textbook_Maps/

Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/ Physical_Properties_of_Matter/States_of_Matter/Properties_of_Gases/Gas_Laws/ The_Ideal_Gas_Law#:~:text=R%20%3D%200.0820574%20L%E2%80%A2atm %E2%80%A2mol%2D1K 3. https://pubchem.ncbi.nlm.nih.gov/compound/Water#section=Chemical-and-Physical- Properties 4. http://butane.chem.uiuc.edu/pshapley/GenChem1/L21/2.html