Due In 2 Hours Quick Please Provide Solutions And Answers

Due In 2 Hours Quick Please Provide Solutions And Answers1

1. What is the molar concentration of NaOH when 23.7786 g is dissolved in enough water to fill a 500.00 mL volumetric flask?

2. What is the molality of a solution prepared by dissolving 12.011 g of sucrose in 250.0 mL of water at 24 °C?

3. A solution is prepared that is 23.44 wt% NaCl in water. What is the molality of the solution?

4. How many grams of PbCl₂ will be produced if excess HCl is added to 30.00 mL of a 0.2319 M solution of Pb(NO₃)₂ according to the following reaction? (Assume that the reaction proceeds to completion. MW PbCl₂ = 278.1164 g/mol.)

5. What is the concentration of a NaOH solution if 34.98 mL of it reacted with 1.9278 g of KHP? (MW KHP = 204.23 g/mol)

Paper For Above instruction

Introduction

This paper provides comprehensive solutions and detailed explanations for a series of chemical problems involving molarity, molality, percent composition, and stoichiometry. These problems are essential in analytical chemistry for understanding solution concentrations and predicting reaction outcomes. Each problem will be addressed systematically to facilitate understanding and to ensure accurate calculations based on stoichiometric principles and chemical formulas.

Problem 1: Molar Concentration of NaOH

The first problem pertains to calculating the molarity of a sodium hydroxide (NaOH) solution. Molarity (M) is defined as moles of solute per liter of solution. Given that 23.7786 g of NaOH is dissolved in water to fill a 500.00 mL volumetric flask, we proceed as follows:

1. Calculate the moles of NaOH: Using the molar mass (MW) of NaOH, which is 39.997 g/mol, the moles are:

   moles = mass / MW = 23.7786 g / 39.997 g/mol ≈ 0.5945 mol

2. Calculate the molarity: Convert 500.00 mL to liters (0.500 L) and use the formula:

   Molarity = moles / volume (L) = 0.5945 mol / 0.500 L ≈ 1.189 mol/L

Thus, the molar concentration of NaOH is approximately 1.189 M.

Problem 2: Molality of Sucrose Solution

Molality (m) is defined as moles of solute per kilogram of solvent. For this problem, 12.011 g of sucrose is dissolved in 250.0 mL of water at 24°C. To calculate molality:

1. Calculate the moles of sucrose: Using MW = 342.30 g/mol,

   moles = 12.011 g / 342.30 g/mol ≈ 0.03506 mol

2. Convert volume of water to mass: Assuming water density at 24°C is approximately 0.997 g/mL,

   mass of water = 250.0 mL × 0.997 g/mL ≈ 249.25 g = 0.24925 kg

3. Calculate molality:

   molality = moles of solute / kg of solvent = 0.03506 mol / 0.24925 kg ≈ 0.1405 mol/kg

Therefore, the molality of the sucrose solution is approximately 0.1405 mol/kg.

Problem 3: Molality of a NaCl Solution

Given that the solution is 23.44 wt% NaCl, this means there are 23.44 g of NaCl in 100 g of solution. To find molality:

1. Determine the mass of water: In 100 g of solution:

   mass of NaCl = 23.44 g

   mass of water = 100 g - 23.44 g = 76.56 g = 0.07656 kg

2. Calculate moles of NaCl: Using MW = 58.443 g/mol,

   moles = 23.44 g / 58.443 g/mol ≈ 0.401 mol

3. Calculate molality:

   molality = 0.401 mol / 0.07656 kg ≈ 5.24 mol/kg

The molality of the NaCl solution is approximately 5.24 mol/kg.

Problem 4: Mass of PbCl₂ Produced

The reaction involves the formation of PbCl₂ from Pb(NO₃)₂ and HCl, where excess HCl ensures complete reaction. Given the volume and molarity of Pb(NO₃)₂ solution:

1. Calculate moles of Pb(NO₃)₂:

   moles = molarity × volume (L) = 0.2319 mol/L × 0.03000 L = 0.006957 mol

2. Determine moles of PbCl₂ formed: From the stoichiometry of the reaction:

   1 mol Pb(NO₃)₂ produces 1 mol PbCl₂.

   moles of PbCl₂ = 0.006957 mol

3. Calculate the mass of PbCl₂:

   mass = moles × MW = 0.006957 mol × 278.1164 g/mol ≈ 1.935 g

The mass of PbCl₂ produced is approximately 1.935 grams.

Problem 5: Concentration of NaOH

Given that 34.98 mL of NaOH reacts with 1.9278 g of KHP, we determine the molarity of NaOH:

1. Calculate moles of KHP: Using MW = 204.23 g/mol,

   moles = 1.9278 g / 204.23 g/mol ≈ 0.00943 mol

2. Determine moles of NaOH: The balanced reaction:

   NaOH + KHP → NaKP + H₂O

   shows a 1:1 molar ratio.

   Thus, moles of NaOH = moles of KHP = 0.00943 mol.

3. Calculate molarity of NaOH: Convert 34.98 mL to liters (0.03498 L) and use:

   Molarity = moles / volume (L) = 0.00943 mol / 0.03498 L ≈ 0.2697 mol/L

The concentration of the NaOH solution is approximately 0.2697 M.

Conclusion

These solutions showcase fundamental concepts of solution chemistry, including molarity, molality, percent composition, and stoichiometric calculations. Accurate application of molar masses, unit conversions, and stoichiometry principles allows for precise determination of solution concentrations and reaction yields. These calculations are integral for laboratory techniques, quality control, and research involving chemical solutions. Mastery of these concepts enhances the ability to analyze and interpret chemical data effectively.

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