CHEM REVIEWER βΊ MOLARITY unit: M or mol/L ο Molar solution of a substance is defined as a solution contains 1g molecular weight (1 mole of the solute in 1L solution) of the substance/liter of the solution ο π= 49π = 1.0πππ/ππ (98 π/πππ)(0.5ππ) π= πππππ ππ π πππ’π‘π πΏ π πππ’π‘πππ πππππππ‘π¦ = πππππ ππ π πππ’π‘π since πππππ = ππππππ’πππ π€πππβπ‘ πππππ ππ π πππ’π‘π (ππ ππ π πππ’π‘π)(πΏ ππ π πππ’π‘π) Example: 1. Calculate the molarity of 2.0 L of solution containing 5.0 moles of NaOH. π= 5.0πππ π πππ’π‘π = 2.5π ππ 2.5πππ/πΏ 2.0πΏ π πππ’π‘πππ βΊ SOLUTIONS ο Homogenous mixtures of 2 or more substances ο Parts: 1) Solute – substance that dissolves in a solution 2) Solvent – dissolving medium βΊ MOLE FRACTION ο The number of moles of 1 component to the total number of moles of all components present ππππ πΉππππ‘πππ ππ ππππ’π‘π = ππππ ππ π πππ’π‘π πππππ ππ π πππ’π‘π + πππππ ππ π πππ£πππ‘ ππππ πΉππππ‘πππ ππ ππππ£πππ‘ = ππππ ππ π πππ£πππ‘ πππππ ππ π πππ’π‘π + πππππ ππ π πππ£πππ‘ 2. Calculate the volume of a 0.750M sulfuric acid (H2SO4) solution containing 0.120 moles of solute. 0.120πππ 0.750π = π₯πΏ π₯ = 0.160πΏ Activity: Example: 1. Calculate the molarity of a solution containing 10g of sulfuric acid (MW = 98) in 500mL of (pakidugtungan kulang sakin hehe) Given: ο 10g sulfuric acid 10π π = (98)(0.5πΏ) = 0.204πππ/πΏ ο 98 MW ο 500mL πππππ π»2π = 2. Calculate the weight in grams of sulfuric acid in 2L of 0.100M solution Given: ο 0.100M solution πππππ = (π)(ππ)(πΏ) ο 2L πππππ = (0.100)(98)(2πΏ) ο 98 MW πππππ = 19.6π βΊ MOLALITY π= 1. What are the mole fractions of solute and solvent in a solution prepared by dissolving 98g sulfuric acid in 162 water? 98π πππππ π»2ππ4 = = 1πππ (π πππ’π‘π) 98π/πππ unit: m or mpl/kg πππ π πππ’π‘π (ππ)(ππ π πππ£πππ‘) Example: 1. What is the molality of a solution in which 49g sulfuric acid is dissolved in 500g H20? 162π = 9 πππ (π πππ£πππ‘) 18π/πππ ππππ πππππ‘πππ π»2ππ4 = 1 = 0.1 1+9 ππππ πππππ‘πππ π»2ππ4 = 9 = 0.9 1+9 βΊ NORMALITY FRACTION ο Gram equivalent weight of solute per liter solution Gram equivalent = π= MW π£ππππππ πππππ π πππ’π‘π (πΈπ)(πΏππ‘πππ ππππ’π‘πππ) Example: 2) Volume in volume (v/v) percent solution 1. Calculate the normality of a solution containing 2.45g sulfuric acid in 2L solution π= 2.45π = 0.025 π ππ πππ’ππ£/πΏ 98 ∗ 2πΏ 2 βΊ PERCENT SOLUTION ο Amount of solute in a solution can be measured as a percentage of total value of the solute. The term percent refers to parts of solute per 100 parts of the solvent 1) Weight in volume (w/v) percent solution ο Volume = 100 π€ πππππ π πππ’π‘π % = π₯ 100 π£ ππ ππ π πππ’π‘πππ ο Or grams of solute per 100ml of solution Example: a) What is the w/v% of solution that has 25.0g of NaCl dissolved into a total volume of %100ml? % π€ 25.0π πππΆπ = π₯ 100 = 25% π£ 100ππ π πππ’π‘πππ ο Therefore, a solution with 25.0g of NaCl dissolved into 100ml of H20 has a %w/v concentration of 25% b) Prepare 200ml of 10% NaOH (10)(200) 100 = 20π ππππ» ππ’πππ‘ππ‘π¦ π π’ππππππππ‘ π‘π 200ππ Activity: a) How many grams of NaOH would be needed to make a 40% w/v solution using deionized water as the solvent? (40π)(100) = 40π 100 b) How many grams of CaCl2 would be needed to make 10% w/v solution using deionized water as the solvent? (10)(100) = 10π 100 % π£ ππ ππ π πππ’π‘π = π₯ 100 π£ 100ππ ο Or ml of solute per 100ml of solution ο % v/v is similar to % w/w in that the total volume of the solution is 100ml ο ππππ’ππ‘ ππ π πππ£πππ‘ = 100ππ π‘ππ‘ππ π£πππ’ππ − ππππ’ππ‘ ππ π πππ’π‘π Example: a) Prepare 200ml of 5% ammonium hydroxide solution π£ ππ π πππ’π‘π % = π₯ 100 π£ 100ππ ππ π πππ’π‘πππ ππ ππ π πππ’π‘π = π₯= %π πππ’π‘πππ πππ ππππ π₯ π‘ππ‘ππ π£πππ’ππ πππ ππππ 100 (5)(200) = 10ππ 100 ο 10ml ammonium hydroxide quantity sufficient to 200ml b) How many ml of ethanol are needed to make a 75% v/v solution using 100ml deionized water as solute? (75)(100) = 75 100 c) How many ml of water are necessary for the solution? π₯ = 100 − 75 π₯ = 25 d) How many ml of HCl are needed to make a 10% v/v solution using deionized water as the solvent? (10)(100) = 10 10 3) Weight in weight percent solution (w/w) % π€ πππππ ππ π πππ’π‘π = π₯ 100 π€ 100π ππ π πππ’π‘πππ Example: a) Prepare 100ml of 10% ammonium hydroxide solution from the concentrated reagents in which 1ml of NaOH contains 3.97g pure ammonia. (3.97)(10)(100) π₯= = 39.7ππ 100 ο 39.7ml concentrated ammonium hydroxide quantity sufficient to 100ml Activity: a) How many grams of NaOH are needed to make a 30% w/w solution using deionized water as the solvent? Median ο Middle value when data are ordered from the smallest ο To determine: Arrange data in order, from smallest to the highest value. If there is an even number of items in the data set, then the median is found by taking the mean (average) of the two middlemost numbers. Mode π€ (30π)(100ππ) 30.0% = 30.0πππππ» π€ 100ππ ο 30.0g would be dissolved into 70.0gH2O βΊ CONCENTRATION CALCULATIONS ο Changing concentrations between 2 solutions πΆ1π1 = πΆ2π2 Where: C1 = stock concentration C2 = concentration of new solution V1 = volume of stock required V2 = volume of new solution ο The most frequently appearing value in the data. Measures of Dispersion ο Describes how widely dispersed the data in the data set. Range ο The difference between the lowest and highest values. In {4, 6, 9, 3, 7} the lowest value is 3, and the highest is 9, so the range is 9 − 3 = 6. Range can also mean all the output values of a function. Standard Deviation ο Describes the spread of individual measurement about the mean. Example: a) 20ml of 2.00M solution are diluted to 100ml in a 100ml volumetric flask. What is the concentration of the new solution? πΆ1π1 (2.00π)(20ππ) πΆ2 = = = 0.4π π2 100ππ b) If there are 100ml of 6.00m NaOH solution but a 5.00M solution a dilute. How much water should be added to the 6.00M solution to dilute it to 5.00M? πΆ1π1 (6.00π)(100ππ) π2 = = = 120ππ πΆ2 5.00π ο CENTRAL TENDENCY ο Quantity that describes the location of a center of data. Mean ο Numerical average by obtaining by dividing the sum of the individual measurements. ο The mean is the average of the numbers. It is easy to calculate: add up all the numbers, then divide by how many numbers there are. In other words it is the sum divided by the count. Relative Standard Deviation ο Divide the standard deviation by the mean. Percent Relative Standard Deviation. ο Multiply the relative standard deviation by 100% Variance ο Square of the standard deviation. ο ACCURACY AND PRESCISION Molality Accuracy ο Closeness of the measurement to the TRUE or ACCEPTED VALUE. ο Usually express as either absolute error or relative error. Absolute Error ο The difference between experimental result and the true value. ο (x) = experimental result ο (µ) = true value ο Formula: e = x - µ Normality Relative Error ο Difference between the true value and measured values and dividing this difference by the true value. (ito definition sa ppt ni sir) Temperature Precision ο Measures the spread of data about a central data. ο Repeatability: is the precision obtained when all measurements are made by the same analyst during a single period of laboratory work. ο Reproducibility: is the precision under any other set of condition, including that between analyst or between laboratory sessions for a single analyst. ο OTHER EXAMPLES Molarity