Can you answer questions 19-22?

I'll give you the conceptual process to answer #19#, but I'll leave it as an exercise for you to get to the answers themselves. Since the remaining questions are short, I'll answer them anyway, but I'll let you fill in some gaps.

#19)#

#a)#

Mass percent is #"g solute"/"g solution" xx 100%#. Assume the density of the solution is that of water, #"1.00 g/mL"#, and so you should have #"153 g solution"# and #"5.00 g solute"#.

#b)#

#"Molarity"# is #"mols solute"/"L solution"#.

Convert the mass of solute (#"C"_11"H"_22"O"_11#) into #"mols"# of it using the given molar mass, then divide by the volume of solution (not water) in #"L"# (#"0.153 L"#).

#c)#

#"Molality"# is #"mols solute"/"kg solvent"#.

Convert the mass of solute (#"C"_11"H"_22"O"_11#) into #"mols"# of it using the given molar mass, then divide by the mass of solvent (water) in #"kg"#. Again, use the density of water as #"1.00 g/mL"#, and recall that #1000# of something is in #1# kilo of it.

#20)#

Well, write the reaction of it dissociating in water.

What you do is identify the common charge of one of the ions, and use conservation of charge to predict the other element's charge. Then, if you count the atoms in the compound, you can then predict how this could turn out:

#"CaCl"_2(aq) -> "Ca"^(2+)(aq) + 2"Cl"^(-)(aq)#

I'll let you count how many particles this is per formula unit.

So then, for #"1 mol"# of #"CaCl"_2#, you'll have #bb"1 mol"# times as many particles around. Simply multiply what you get right above by #"1 mol"#.

#21)#

Here, you apply the same concept as in question #20# to count how many #bb("NO"_3^(-))# ions you get per formula unit. That is then going to give you an integer multiple of the given concentration.

As an example,

#"Ca"("NO"_3)_2(aq) -> "Ca"^(2+)(aq) + 2"NO"_3^(-)(aq)#

#=> "0.100 M Ca"("NO"_3)_2 -> "0.200 M NO"_3^(-)#,

for #2# #"NO"_3^(-)# particles per #"Ca"("NO"_3)_2# formula unit

Now I'll let you do the same thing for the other three compounds to determine which solution it is that contains the largest concentration of #"NO"_3^(-)#.

[Hint: it is the one that is #"0.225 M"# in #"NO"_3^(-)#.]

#22)#

There is almost no difference. The only difference in principle is the direction of heat flow.

• Melting point defines what temperature something changes from a solid to liquid due to adding heat.

• Freezing point defines what temperature something changes from a liquid to solid due to taking heat away.

Otherwise, the temperature value is the same for both. In practice, they are the same thing.