Ana chem - Lecture notes 2 PDF

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M2 Activity 3Solve the following problems. Use intermediate pad paper, one for Part A and another sheet for Part B. Write legibly and no erasures. Follow the given format below. Final answers must contain the correct number of significant figures. Upload your assignment as a picture.Part A. Problem ...

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M2 Activity 3 Solve the following problems. Use intermediate pad paper, one for Part A and another sheet for Part B. Write legibly and no erasures. Follow the given format below. Final answers must contain the correct number of significant figures. Upload your assignment as a picture. Part A. Problem Solving on Molarity 1. 2. 3.

A 200. mL solution contains 4.00 g Ba(OH)2. Express the concentration of the solution in M. How many millilitres of 0.25 M sucrose solution contains 50.0 mg of the solute? A solution was prepared by dissolving 6.34 g KCl.MgCl2.6H2O in sufficient water to give 2.00 L. Calculate: (a) M of KCl.MgCl2.6H2O

4.

(b) M of KCl.MgCl2

(c) M of Mg+2

(d) %w/v KCl.MgCl2.6H2O

A 15.0 % (w/w) NiCl2 solution has a density of 1.149 g/mL. Calculate the Molar concentration of 1L of this solution.

Part A. Problem Solving on Molarity Problem # Formula 1.Given: Molarity= Moles of solute Liters of solution = Liters of solution 200. mL0.2 L Moles of solute = 4.00 g Ba(OH)2 Req’d: M? 2.Given: Molarity= Moles of solute Molarity: 0.25 M Liters of solution Mole of solute:50.0 mg Req’d: mL=?

Solution M= 4.00 g Ba(OH)2 X 1 mol Ba(OH)2 X 1 1 171.34 g Ba(OH)2 0.2 L M= 0.12 mol/ L

0.05gX 1 mol C12H22O11= 1.46 mol 342 g

3.A. Given: Moles of solute = 6.34 g KCl.MgCl2.6H2O Liters of solution = 2.00L Req’d: M of KCl.MgCl2.6H2O

Molarity= Moles of solute Liters of solution

0.25M X 1.46 mol = 5.84L X1000=5840 mL 0.25M 0.25 mol/L 6.34g KCl.MgCl2.6H2O X 1 mol = 0.01mol/L 2.00L 277.85g

3.B. Given: Moles of solute = 6.34 g KCl.MgCl2.6H2O Liters of solution = 2.00L Req’d: M of KCl.MgCl2 3.C. Given: Moles of solute = 6.34 g KCl.MgCl2.6H2O Liters of solution = 2.00L Req’d: M of Mg+2 3.D. Given: Mass of solute = 6.34 g KCl.MgCl2.6H2O Volume of solution = 2.00L  2000mL Req’d: %w/v KCl.MgCl2.6H2O

Molarity= Moles of solute Liters of solution

6.34g KCl.MgCl2.6H2O X 1mol KCl.MgCl2 = 0.02mol/L 2.00L 169.762g/mol KCl.MgCl2

Molarity= Moles of solute Liters of solution

6.34g KCl.MgCl2.6H2O X 1mol KCl.MgCl2 = 0.13mol/L 2.00L 24.305 g/mol Mg

w/v% = mass solute × 100 Volume of solution

w/v% = 6.34 g KCl.MgCl2.6H2O X100= 0.32 % 2000mL

Part B. Problem Solving on Normality 1.

A 200. mL solution contains 4.00 g Ba(OH)2. Express the concentration of the solution in Normality. How does it compare with the solution’s Molarity?

2.

3. 4.

Calculate the equivalent weights of the following acids assuming that all the H+ each contains are involved in the reactions: (a) HCl (b) H2SO4 (c) H3PO4 How many grams of pure Na2CO3 is needed to prepare 500. mL of 0.1000 N Na2CO3 solution? What is the Normality of KMnO4 solution if 500. mL of it contains 5.00 g of the solute? ( Mn +7 is reduced to Mn+2).

Problem # 1.Given: Liters of solution = 200. mL0.2 L Moles of solute = 4.00 g Ba(OH)2 Gram eq of solute= 4.00 g Ba(OH)2 Volume of sol. In liter= 200. mL  0.2 L

Formula Molarity= Moles of solute Liters of solution Normality= Gram eq of solute Volume of sol. In liter

Solution M= 4.00g X 1 mol = 0.12 0.2L 171.34 g/mol Gram eq of solute = 1mol H+ X 1 mol Ba(OH)2 X 171.34 g/mol = 85.67g 1 2 mol H+ 1 mol Ba(OH)2

Normality= 4.00g X 1 ew X 1 85.67g

1 0.2 L

= 0.23Eq/L

Req’d: Molarity and normality=? 2.a. HCL Given: Molecular weight= 36.461 g/mol Valency= +1 Req’d= E. W=? 2. b. H2SO4 Given: Molecular weight= 98.18 g/mol Valency= +2 Req’d= E. W=? 2c. H3PO4 Given: Molecular weight= 97.99g/mol Valency= +3 Req’d= E. W=? 3. Given: Normality=0.1000 N Na2CO3 Volume of sol. In liter=500. mL  0.5 L Req’d= grams=?

4.Given: Liters of solution = 500. mL0.5 L Moles of solute = 5.00 g Req’d= Normality of KMnO4

E.w= 36.461 g/mol /1 = 36.461 Equivalent Weight = Molecular weight/ Valency E.w= 98.18 g/mol / 2 = 49.09

Equivalent Weight = Molecular weight/ Valency E.w= 97.99 g/mol /3 = 32.66 Equivalent Weight = Molecular weight/ Valency Normality= Gram eq of solute Volume of sol. In liter

E.w= 105.9888 g/mol/ 2 = 52.9944 0.1000 N Na2CO3 X52.9944 X 0.5 = 0.5 g

Molarity= Moles of solute Liters of solution Normality= Gram eq of solute Volume of sol. In liter

M= 5.00g X 1 mol = 0.06 0.5L 158.034 g/mol Normality= 0.06 X 2= 0.12 Eq/L

M2 Activity 4 A HNO3 solution has a sp.gr. of 1.42 and is 70.0 % (w/w) HNO 3. Express its concentration in (a) molality (b) in Molarity (c) how many ml of this solution is needed to prepare 250. mL of 1.00 M HNO 3? (d) how many mL of 1.00 M NaOH is needed to neutralize 25.0 ml of 1.00 M HNO3? Problem solving on Molality and Dilution and Neutralization Problem

Problem # a.Given: sp.gr. of 1.42 70.0 % (w/w) Req’d: 1. Density of solution= ? 2. Mass of solution =? 3. Mass of solute =? 4. Mass of solvent=? 5. Molarity of solution=?

Formula Molarity= Moles of solute Liters of solution

Solution S.G=ρHNO3ρwater⇒ρHNO3=S.G.⋅ρwater ρHNO3=1.42⋅1 g/mL=1.42 g/mL 1.0L⋅1000mL⋅1.42 g1mL=1420 g 1L 1mL 1420g solution⋅ 70 g HNO3 =994 g HNO3 100g solution 994g⋅1mole=15.78 moles HNO3 63.01g C=n=15.78 moles =15.78 mol/L V 1.0 L

b. Given: sp.gr. of 1.42 70.0 % (w/w) Molarity= 15.78 mol/L Req’d : Molality =?

Molality= moles of solute Kg of solvent

c. Given: 70.0 % (w/w) No. of mole=0.25 mole 1.00M HNO3 Req’d : mL=? d.Given: 1.00 M NaOH 25.0 ml 1.00 M HNO3

Mass= No. of moles X molecular mass

No of moles: 1.00M X 0.25 L=0.25 mole Mass= 0.25 mole X 63.01 g/mol= 15.75 g/mole 70% w/w= 15.75 X 100 =22.5 mL 70

molarity=mol solute L soln

HNO3+NaOH→NaNO3+H2O Mol HNO3=1.00 X 0.025L = 0.025 mol HNO3 0.025 mol HNO3(1 mol NaOH)= 0.025 mol NaOH 1 molHNO3 0.025 mol NaOH = 0.03 L= 30mL 1.00

m=1420−994=426 g water b= 15.78 moles=37.04 mol/Kg moles426⋅10−3kg

Req’d : mL=?

M6 Pre-Task How well do you recall previous lessons on naming and writing formula of common anions? For nos. 1-3 given the name of the anion, write its formula and for nos. 4-6 given the formula write its name. Type your answers in the reply section. Make positive comments on your classmate answers.

1. 2. 3.

acetate- C2H3O2sulfate- SO₄²chromate-CrO₄-2

4. 5. 6.

NO3-1 -Nitrate ion C2O4-2 -Oxalate ClO3-1 -Chlorate

potassium dichromate (the chromium is reduced to Lecture: M2 Lesson 3 Normality chromium +3) 3. NORMALITY (N) weight M2 Lesson 4 Molality -------------EW wt 1000 mL/L N = ------------------ = ----------- x ------------------4. MOLALITY (m) L solution EW mL weight solution ------------MW wt 1000 Unit: eq g/kg ---------m = ------------------- = ---------- x -----------------Liter Kg solvent MW g solvent MW g/mole g EW = ---------- = ----------------- = Unit: mole ---------------------kilogram # (--) eq/mole eq Acid : # H+ Base: # OH Situations in which the properties of the Salt: # cation solvent are studied. 

Used to express concentrations of 1) Acids and bases 2) Oxidizing agents 3) Reducing agents N =M x n Sample Problems 1. Calculate the equivalent weights of the following substances: (a) ammonia, (b) oxalic acid (in reaction with sodium hydroxide), (c) potassium permanganate (manganese 7 is reduced to manganese 2) 2. Calculate the normality of the solutions containing the following: (a) 5.300 g/L (when the carbonate reacts with two protons), (b) 5.267 g/L

Sample Problem 1. How many milliliters of concentrated sulfuric acid, 94.0% (g/100 g solution), density 1.831 g/cm3, are required to prepare 1 liter of a 0.100 M solution? Read the article and watch the videos below to further understand the lesson. M2 Lesson 6 Parts Per Million and Parts Per Billion

6. PARTS PER MILLION (ppm) gram of solute ppm = ----------------------------- x 106 gram of solution

ppm = -----------------------------------kilogram of solution

milligram of solute ppm = ----------------------------------Liter of solution 

milligram of solute

It is often used to express the concentration of very dilute solutions.

PARTS PER BILLION (ppb) gram of solute ppb = ----------------------------- x 109 gram of solution microgram of solute ppb = -------------------------------------Liter of solution microgram of solute ppb = --------------------------------------kilogram of solution



It is also used to express the concentration of very dilute solutions.

1ppm = 1000ppb Sample Problems 1. A 2.6 g sample of plant tissue was analyzed and found to contain 3.6 microgram zinc. What is the concentration of zinc in the plant in ppm? In ppb? 2. A 25.0 microliter serum was analyzed for glucose content and found to contain 26.7 micrograms. Calculate the concentration of glucose in ppm and in mg/dL. 3. The concentration of zinc ion in blood serum is about 1 ppm. Express this in meq/L....