7.34, 7.42, 7.50, 7.62, 7.66, 8.24, 8.42, 8.52, 8.56, 8.80

7.34 - The conversion factors between kPa (kilopascals) and the other units involved are:

1 atm = 101.3 kPa = 760 mm Hg

(a) To convert from kPa to atm:

(b) to convert from kPa to mm Hg

7.42    To solve any calculation problem like this one, there is a few steps that you can take to make it easy. 

First, write out what is given (either explicit information written in the problem or equations you know will help you in the problem) and what needs to be found.

 Find: ? L of NH₃

 Given:  3H₂(g) + N₂(g) --> 2NH₃(g)

0.75 mol of H₂, which we can assume reacts completely with N₂ to form ammonia.

Next, convert to moles (which doesn’t have to be done in this problem since the initial substance is given in moles) and then use conversion factors (from the coefficients of the balanced reaction) to find the desired answer.     The problem asks for the answer in liters     

Next, convert from moles to the desired quantity of measure which in this case is Liters of NH₃

 The problem says that we are at standard temperature and pressure.  At STP,

1 mol = 22.4 L (from Avogadro’s hypothesis and molar volume of gas at STP p. 185).  Using

we can conversion factor to calculate volume in liters.

7.50    To solve, two principles need to be understood--first, the effect on the pressure after changing the size of the container (volume) that contains the gas (p. 190) and secondly, Dalton’s law of partial pressure (p. 192).

 First, we can consider each gas independently.  Each gas is forced from a 2 L container to  a 1 L container.  This decrease in size creates more collisions between the gas molecules.  Since the volume is reduced by ½, the pressure of each gas doubles.  Therefore, the partial pressure of each gas will increase from 1 atm pressure to 2 atm pressure.  This follows from the relation

P₁V₁ = P₂V₂

For each of the gases, P₁ = 1 atm, V₁ = 2L and V₂ = 1L

Solving for P₂ and plugging in yields

Hence each gas will have a partial pressure of 2 atm in the new container.

Second, using Dalton’s law of partial pressure, the total pressure will be the sum of the partial pressures of the gases.  Hence, there are three gases each with a partial pressure of 2 atm, so the total pressure is P_tot = 2+2+2=6 atm.

7.62    First, write out what is given (either explicit information written in the problem or equations you know will help you in the problem) and what needs to be found.

 Given:  (for all three parts)

PV = nRT (ideal gas law, p.201)

R = 0.0821 (L x atm)/(K x mol) (Ideal gas constant)

The conditions of STP are 1atm and 273K.  At STP, one mole of an ideal gas          occupies 22.4 L.

Find: ? L of gas

We are given either moles or mass in this problem.  We have a relation between moles and volume so we will need to work with moles.  In (b), we need to use the molar mass to convert from mass to moles.  

Once we have moles, there are two ways to do each of these calculations.  The first is to simply use the ideal gas law.  The question asks for volume so if we solve the ideal gas law for volume and plugging in the conditions of STP we get

The second is simply to use the molar volume of gas at STP which we know is 22.4 L / mol.  The second method is easier for this problem, but remember it can only be used at STP.  

(a) Find: V = ?L H₂

 Given:  n = 6.20 mol H₂

P = 1 atm (conditions are at STP)

T = 273 K (conditions are at STP)

            R = 0.0821 L atm mol^-1 K^-1

            Solve: V = (n  R  T)/ P

V = (6.20 mol H₂ x 0.0821 L atm mol^-1 K^-1  x 273 K) / 1 atm = 139 L H₂

Alternatively, we simply use the molar volume at STP: 

V = (n)(22.4 L mol^-1 ) = (6.20 mol)(22.4 L mol^-1) = 139 L H₂

(b) Find: V = ?L N₂

            Given: 6.8 g N₂             M.W. N₂ = 28 g/mol

n = ? mol

            P = 1 atm (conditions are at STP)

            T = 273 K (conditions are at STP)

                        R = 0.0821  L atm mol^-1 K^-1

           Solve:

First, convert g to mol of N₂.

 6.8 g N₂ * mol N₂ / 28 g = 0.2429 mol N₂ = n

Second, plug correct values into the ideal gas law and solve.

V = (n x R x T)/ P

V = (0.243 mol N₂ x 0.0821 L atm mol^-1 K^-1 x 273 K) / 1 atm = 5.4 L N₂

or

V = (n)(22.4 L mol^-1 ) = (0.243 mol)(22.4 L mol^-1) = 5.4 L N₂

(c) Find: V = ?L CO₂

             Given:  n = 2.35 mol CO₂

            P = 1 atm (conditions are at STP)

            T = 273 K (conditions are at STP)

                        R = 0.0821L atm mol^-1 K^-1

           Solve: V = (n x R x T)/ P

V = (2.35 mol CO₂ x .0821 (L x atm)/(K x mol) x 273 K) / 1 atm = 52.6 L CO₂

           or

        V = (n)(22.4 L mol^-1 ) = (2.35 mol)(22.4 L mol^-1) = 52.6 L CO₂

7.66 - This problem asks us to interrelate the pressure, temperature and volume of a given number of moles of gas confined to a balloon.  Since none of P, T or V are held constant, we need to use the combined gas law:

P₁V₁ / T₁ = P₂V₂ / T₂

We are given P₁ = 1.00 atm, T₁= 27^oC = 300K, V₁ = 66L, P₂ = 380 mm Hg (1 atm / 760 mm Hg) =  0.5 atm, T₂ = -23^oC = 250 K.  Solving for V₂ and plugging in yields,

V₂ = P₁V₁T₂ / (T₁P₂) = (1.00 atm)(66L)(250K) / [(300K)(0.50 atm)] = 110 L.

8.24 – The boiling point of a substance is related to the strength with which the molecules of the substance are held together.  The strong these intermolecular interactions, the higher the boiling point because it takes more energy to break molecules of the liquid up.  Water molecules are held together by a relatively strong type of intermolecular interaction termed hydrogen bonding.  Consequently, it has a unexpectedly high boiling point relative to similar compounds that do not exhibit hydrogen bonding.   

8.42 – The solubility of NaCl is given as 36.0 g / 100 mL = 0.36 g / mL.  If there is more NaCl dissolved then the solution is supersaturated.  If there is precisely 0.36 g / mL of NaCl dissolved then the solution is saturated.  If there is less than 0.36 g / mL of NaCl dissolved then the solution is unsaturated.  The problem inquires about a solution containing 26.5 g NaCl / 75.0mL H₂O  = 0.35 g / mL, which is less than 0.36 g / mL so the solution is unsaturated.

8.52 – 

(a) One part in a million refers to one piece of a million.  In terms of mass, we can consider 1 ppm as 1 g in 106g.  For a solution in water, 1 g of solution is approximately 1mL so

This is often expressed in units of mg / L

So 1ppm = 1 mg / L. 

If we convert, the concentration given to mg / L we will have ppm.  The problem states 0.00024 g / 100mL

 (b) Molarity is mol / L.  To find mol we need the molar mass of BaSO₄ = 233.4 g / mol

  8.56

The molarity and volume will allow us to calculate moles.  To convert from moles to grams, we use the molar mass.

  (a) Molar mass of NaCl = 58.5 g / mol

  (b) Molar mass of KMnO₄  = 158.0 g / mol

(c) Molar mass of CaCl₂ = 111.1 g / mol

(d) Molar mass of AgNO₃ = 169.9 g / mol

8.80 

(a) Dissolution requires solvation of sugar molecules at the surface of the solid.  The more surface area available for this process, the faster the dissolution.  If you think of dissolution as a chemical reaction, we have previously learned that chemical reactions involving solids are faster the greater the surface area.

(b) The solubility of gases in liquids increases as the temperature decreases.  Consequently, there will be more oxygen available in the colder bowl.

(c) CaSO₄ can accommodate two waters of crystallization.  Consequently, it can absorb water from the atmosphere.

(d) If we add sugar until no more will dissolve, the solution is saturated.  The fact that solid sugar exists at the bottom of the cup before we add more sugar tells us that we are starting with a saturated solution (otherwise the sugar crystals would dissolve).  As the solution is already saturated, the added sugar will simply settle to the bottom without further dissolution so the tea will taste no sweeter.

(e) Water is polar and dissolves other polar substances.  As oil is nonpolar, it does not dissolve in water resulting in two