What mass of solute is required to prepare a #0.875*"molal"# solution from #534*mL# of water solvent?

1 Answer
Aug 12, 2016

Answer:

Approx. #27*g#. I assume you mean a #"molal concentration"#.

Explanation:

#"Molality"="moles of solute"/"kg of solvent"#

Thus #0.875*mol*kg^-1# #=# #"number of moles of NaCl"/(0.534*kg)#

And #"number of moles of NaCl"# #=# #0.875*mol*kg^-1xx0.534*kg# #=# #0.467*mol#

#"Grams of NaCl"=0.467*molxx58.44*g*mol^-1#. The #"molar concentration"# would be to all intents and purposes identical.