For diluting a 10.0 M solution to a 1.00 M solution whether the instruction: “take 100.0 mL of the 10.0 M solution and add 900.0 mL water” is correct or not is to be explained. Concept introduction: Molarity ( M ) is one of the concentration terms that determine the number of moles of solute present in per liter of solution. Unit of molarity is mol/L . The expression to calculate the molarity of a solution when moles of solute and volume of solution are given is as follows: Molarity of solution ( M ) = moles of solute ( mol ) volume of solution ( L ) Dilution is the process of converting a concentrated solution into a dilute solution by adding more solvent. In the resultant solution, the amount of solute remain fixed but the volume of the solution increases resulting in a less concentrated solution.

Question

Want to see the full answer?

Answer 1

Question

Chapter 4, Problem 4.13P

Interpretation Introduction

Interpretation:

For diluting a 10.0 M solution to a 1.00 M solution whether the instruction: “take 100.0 mL of the 10.0 M solution and add 900.0 mL water” is correct or not is to be explained.

Concept introduction:

Molarity (M) is one of the concentration terms that determine the number of moles of solute present in per liter of solution. Unit of molarity is mol/L.

The expression to calculate the molarity of a solution when moles of solute and volume of solution are given is as follows:

Molarity of solution(M)=moles of solute(mol)volume of solution(L)

Dilution is the process of converting a concentrated solution into a dilute solution by adding more solvent. In the resultant solution, the amount of solute remain fixed but the volume of the solution increases resulting in a less concentrated solution.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Students have asked these similar questions

When 15.00 mL of 3.00 M NaOH was mixed in a calorimeter with 12.80 mL of 3.00 M HCl, both initially at room temperature (22.00 C), the temperature increased to 29.30 C. The resultant salt solution had a mass of 27.80 g and a specific heat capacity of 3.74 J/Kg. What is heat capacity of the calorimeter (in J/C)? Note: The molar enthalpy of neutralization per mole of HCl is -55.84 kJ/mol.

When 15.00 mL of 3.00 M NaOH was mixed in a calorimeter with 12.80 mL of 3.00 M HCl, both initially at room temperature (22.00 C), the temperature increased to 29.30 C. The resultant salt solution had a mass of 27.80 g and a specific heat capacity of 3.74 J/Kg. What is heat capacity of the calorimeter (in J/C)? Note: The molar enthalpy of neutralization per mole of HCl is -55.84 kJ/mol. Which experimental number must be initialled by the Lab TA for the first run of Part 1 of the experiment? a) the heat capacity of the calorimeter b) Mass of sample c) Ti d) The molarity of the HCl e) Tf

Predict products for the Following organic rxn/s by writing the structurels of the correct products. Write above the line provided" your answer D2 ①CH3(CH2) 5 CH3 + D₂ (adequate)" + 2 mited) 19 Spark Spark por every item. 4 CH 3 11 3 CH 3 (CH2) 4 C-H + CH3OH CH2 CH3 + CH3 CH2OH 0 CH3 fou + KMnDy→ C43 + 2 KMn Dy→→ C-OH ") 0 C-OH 1110 (4.) 9+3 =C CH3 + HNO 3 0 + Heat> + CH3 C-OH + Heat CH2CH3 - 3 2 + D Heat H 3 CH 3 CH₂ CH₂ C = CH + 2 H₂ → 2 2