What is the molality of a solution containing 30.0 g of naphthalene dissolved in 500.0 g of toluene

Question: How many grams of sodium chloride per kilogram of water are present in a 2.7 m aqueous solution?

1. Write the equation for calculating molality:

   molality = moles(solute) ÷ mass(solvent in kg)

2. Rearrange equation to find moles(solute):

   molality

   =

   moles(solute)   mass(solvent in kg)

   
   

   molality × mass(solvent in kg)

   =

   moles(solute) ×   mass(solvent) mass(solvent)

   
   

   moles(solute)

   =

   molality × mass(solvent in kg)

3. Identify the solute and solvent that make up the solution:

   solute = sodium chloride = NaCl
   solvent = water

4. Calculate moles of solute :

   moles(solute) = molality × mass(solvent in kg)
   molality = 2.7 m mass(solvent in kg) = 1 kg

   moles(NaCl) = 2.7 × 1 = 2.7 mol

5. Calculate mass of solute :

   moles(solute) = mass(solute in g) ÷ molar mass(solute in g mol-1) Rearrange equation to find mass:

   mass(solute) = moles(solute) × molar mass

   moles(NaCl) = 2.7 mol

   molar mass(NaCl) = 22.99 + 35.45 = 58.44 g mol-1

   
   mass(NaCl) = 2.7 × 58.44 = 157.79 g

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What is the molality of a solution containing 30.0 $\mathrm{g}$ of naphthalene $\left(\mathrm{C}_{10} \mathrm{H}_{8}\right)$ dissolved in 500.0 $\mathrm{g}$ of toluene?

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We don’t have your requested question, but here is a suggested video that might help.

What is the molality of a solution containing 30.0 $\mathrm{g}$ of naphthalene $\left(\mathrm{C}_{10} \mathrm{H}_{8}\right)$ dissolved in 500.0 $\mathrm{g}$ of toluene?

Welcome to "Colligative Properties" a section refreshing your memory on Chapter 10, here you will find all you need to help you with boiling point, and freezing point along with using molarity and molality.

 This Chapter introduces new vocabulary that you should be familiar with such as:

1. Vapor pressure: The pressure exerted by a vapor on the container and the liquid beneath it. Vapor pressure is lowered when a solute is added to a solvent.

2. Boiling point: The temperature at which the vapor pressure equals the atmospheric pressure. Boiling point is elevated/raised when a solute is added to a solvent.

3. Freezing point: The temperature at which liquid particles lose sufficient energy to become locked in position. Freezing point is lowered when a solute is added to a solvent.

Here are some example sample problems to help you practice on solution concentrations:1. You have 1500 g bleach solution. The percent by mass of sodium hypochlorite, NaOCl, is 3.62%. How manygrams of NaOCl are in the solution?2. What is the molality of a solution containing 30.0 g of naphthalene (C10H8) dissolved in 500.0 g of toluene?

Answer Key: 1. 3.62% = (3.62 g NaOCl / 100 g solution) ----> ( 3.62 g NaOCl / 100g solution) = (X g NaOCl/ 1500g solution)---> x = 54.3 g NaOCl

2. 30 g C10H8 x ( 1mol / 128.164 g) = 0.234 mol ----> m = (0.234 mol / .500 kg) = .468m

Here is a short video to help explain how to find your freezing and boiling point.

Now lets practice on Boiling Point and Freezing Point: 1.Calculate the boiling point and freezing point of each of the following 55.4 g NaCl and 42.3 g KBr dissolved in 750.3 mL water

Answer Key:

55.4 g NaCl x (1mol NaCl / 58.44 g)  = 0.948 mol NaCl    42.3 g KBr x (1 mol KBr/ 119g ) = 0.355 mol KBr

m = ((0.948 mol + 0.355 mol)/ .7503 kg) = 1.74m           *Since NaCl and KBr each split into 2 ions, i = 4