Answer key for a chemistry table (chapter 12)?

Well, I can provide three of them, but you'll have to work the rest out yourself.

For #(2)#, convert #"mL"# to #"L"#, use the definition of molarity to cancel #"L"# and get #"mol solute"#, and convert from #"mol"# to #"g"# of solute.

For #(3)#, convert #"g"# to #"mol"# and use the definition of molarity to calculate the molarity.

For #(4)#, convert #"g"# to #"mol"# and write it in such a way to cancel out the #"mols"# in the definition of molality to get to #"kg"# of solvent.

For #(7)#, convert #"g"# to #"mol"# and write it in such a way to cancel out the #"mols"# in the definition of molarity to get the #"L"# of solvent.

General definitions you MUST know...

#"Molarity" = "mol solute"/"L solution"#

#"Molality" = "mol solute"/"kg solvent"#

#%"w/w" = "g solute"/"100 g solution"#

#"mass solution = mass solute + mass solvent"#

And of course,

#"1000 g"# #=# #"1 kg"#

#"1000 mL"# #=# #"1 L"#

Taking the #"1.15 molal NH"_4"OH"# in #"1200 g"# solvent, we solve for the mass of solute.

#1200 cancel"g solvent" xx cancel"1 kg"/(1000 cancel"g") xx "1.15 mol solute"/cancel"kg solvent"#

#= 1.3_8# #"mols solute"#

where the subscript indicates the digit past the last significant digit. And thus, knowing the molar mass, the mass of ammonium hydroxide found in this solution is...

#1.3_8 cancel("mols NH"_4"OH") xx "35.046 g"/cancel("1 mol NH"_4"OH") = color(blue)("48 g NH"_4"OH")#

(with the solution presumed aqueous ammonia, if water is the solvent), to two sig figs.

Given #"38.9 g solute"# and #"1.4 kg solvent"# , what better concentration to give than molality? You have no density.

#38.9 cancel("g MgCl"_2) xx "1 mol"/(95.211 cancel("g MgCl"_2)) = "0.408"_6# #"mols"#

And so, the molality of the solution is:

#(0.408_6 "mols solute")/("1.4 kg solvent") = color(blue)("0.29 mol/kg")#

or simply #"0.29 m"#... but I prefer mol/kg so it doesn't look like meters.

From "#20.0%#" #"NH"_4"Cl"# (I'm assuming by mass?), we have #"10. g"# of solute, so...

#=> "20. g solute"/"100 g soln" = "10. g solute"/"50 g soln"#

(where the 100 and 50 g are exact!)

We know that in terms of mass, solution = solute + solvent... so the mass of the solvent is...

#"50 g soln" - "10. g solute" = color(blue)("40. g solvent")#