How would I go about doing this question, can someone explain it step by step please, my answer I keep getting is 1.17m CaCl2, but the answer is 1.35m CaCl2?
A 1.30M solution of CaCl2 in water has a density of 1.1g/mL/ What is the MOLALITY?
Also, when for questions with relation to finding minimum pressure, or boiling point elevation or freezing point deperession quesitons, how do I know that I have to use the "i" factor or not?
Thanks
1.30 M means 1.30 moles / liter of solution
1 mole CaCl2 = 111.0 g (How do you get that?)
1.30 mol x 111 g = 144.3 g CaCl2
1.00 L = 1000 mLs
volume x density = mass:
1000 mLs x 1.1 g/mL = 1100 grams solution (total)
1100 g - 144.3g = 955.7 g H2O ---> 0.9553 kg H2O
molality = m = 1.3 mol CaCl2 / 0.9553 kg H2O = ???
The above answer should be multiplied by 3 since 1 mole of CaCl2 produces THREE particles in solution. The new answer is the EFFECTIVE molality you use to get vapor pressure, melting point lowering , and boiling point rise. You got some studying to do on colligative properties and calculations related to that. We do not teach complete units on a bulletin board posting.
1.30 M = 1.30mols of solute/L of solvent
mass of Cacl2 = 1.30 mols CaCl2 x 111g/mol CaCl2(molar mass)= 144.3g of CaCl2 (solute)
Solution = 1.11g/ml
mass of solution = 1.11g/ml x 1000ml/L x1L
= 1110g (solution)
mass of solvent = 1110g (solution) - 144.3g (solute)= 965.7g of solvent x 1kg/1000g = 0.9657kg
molality = 1.30 mols of solute/0.9657kg solvent
= 1.35m
To calculate the molality of the CaCl2 solution, you can follow these steps:
Step 1: Write down the given information:
- Molarity of the solution (M) = 1.30 M
- Density of the solution (ρ) = 1.1 g/mL
- Molar mass of CaCl2 = 110.98 g/mol
Step 2: Convert the density to grams per liter (g/L):
Since the density is given in grams per milliliter, we need to convert it to grams per liter by multiplying it by 1000.
Density (ρ) = 1.1 g/mL x 1000 mL/L = 1100 g/L
Step 3: Calculate the moles of solute (CaCl2):
Moles of solute = Molarity x Volume of solution (in liters)
The volume is not given, so we cannot calculate the exact number of moles. However, we can assume a convenient volume like 1 liter for calculation purposes.
Moles of solute = 1.30 M x 1 L = 1.30 moles
Step 4: Calculate the mass of the solvent (water):
To determine the mass of the solvent, we need to subtract the mass of the solute from the total mass of the solution.
Mass of solute = moles of solute x molar mass of CaCl2
Mass of solute = 1.30 moles x 110.98 g/mol = 144.27 g
Mass of solvent = Total mass of solution - Mass of solute
Mass of solvent = 1100 g - 144.27 g = 955.73 g
Step 5: Calculate the molality (m):
Molality (m) = Moles of solute / Mass of solvent (in kg)
Convert the mass of solvent from grams to kilograms by dividing by 1000.
Molality (m) = 1.30 moles / (955.73 g / 1000) = 1.30 / 0.956 = 1.36 m
So, the correct molality of the CaCl2 solution is 1.36 m, not 1.17 m or 1.35 m as mentioned in your question.
Now, onto your second question about using the "i" factor in questions related to minimum pressure, boiling point elevation, or freezing point depression.
In colligative properties, such as boiling point elevation and freezing point depression, the number of particles (ions or molecules) affects the extent of the property change. The "i" factor (van 't Hoff factor) takes into account the dissociation or association of solute particles in the solution.
To determine whether to use the "i" factor, you need to consider the solute's properties:
1. For non-electrolytes: "i" factor is equal to 1. Non-electrolytes do not dissociate or form ion pairs in solution, so the number of particles doesn't change.
2. For electrolytes: "i" factor is greater than 1. Electrolytes ionize or dissociate in solution, increasing the number of particles.
It is essential to check if the solute in your question is an electrolyte or non-electrolyte. If it is an electrolyte, you will need to use the appropriate "i" factor when calculating the extent of the property change.
Hope this helps! Let me know if you have any further questions.
To solve the first question, you want to calculate the molality of the CaCl2 solution. Molality (m) is defined as the number of moles of solute per kilogram of solvent. Here are the steps to solve it:
Step 1: Convert the given density to kg/mL.
Density = Mass (g) / Volume (mL)
Given: Density = 1.1 g/mL
To convert it to kg/mL, divide by 1000:
Density = 1.1 g/mL / 1000 = 0.0011 kg/mL.
Step 2: Calculate the mass of the solvent.
Since you have the density and the volume of the solution, you can calculate the mass of the solution.
Given: Volume of solution = 1 L (1000 mL)
Mass = Volume × Density = 1000 mL × 0.0011 kg/mL = 1.1 kg.
Step 3: Calculate the number of moles of CaCl2.
To find the number of moles, you need to know the molar mass of CaCl2. The molar mass of Ca = 40.08 g/mol, and the molar mass of Cl = 35.45 g/mol.
Given: Concentration of CaCl2 = 1.30 M
Molar mass of CaCl2 = (40.08 g/mol + 2 × 35.45 g/mol) = 110.98 g/mol
Number of moles = Concentration × Volume = 1.30 mol/L × 1 L = 1.30 mol.
Step 4: Calculate molality.
Molality (m) = moles of solute / kilograms of solvent
Given: Mass of solvent = 1.1 kg
Molality = 1.30 mol / 1.1 kg = 1.18 mol/kg.
The correct answer for the molality is found to be 1.18 mol/kg, not 1.35 mol/kg.
Regarding your second question about the "i" factor, it refers to the Van't Hoff factor. The Van't Hoff factor (i) takes into account the number of particles the solute dissociates into when it dissolves in a solvent. When dealing with non-electrolytes (substances that do not dissociate into ions when dissolved), the "i" factor is equal to 1. However, for electrolytes (substances that dissociate into ions when dissolved), the "i" factor will be greater than 1.
For calculations involving colligative properties like minimum pressure, boiling point elevation, or freezing point depression, you would use the "i" factor when you have an electrolyte involved. It helps adjust the final values based on the extent of dissociation of the solute.