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

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Answer 1 · 2016-08-12T18:35:57

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

#"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.

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