What mass of solute is required to prepare a #0.875*"molal"# solution from #534*mL# of water solvent? Chemistry Solutions Molality 1 Answer anor277 Aug 12, 2016 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. Answer link Related questions How do molality and molarity differ? How high can molality be? How do you calculate molality from molarity? Why is molality used for colligative properties? Why is molality independent of temperature? How can I calculate the molality when 75.0 grams of MgCl2 is dissolved in 500.0 g of solvent? 49.8 grams of KI is dissolved in 1.00 kg of solvent. What is the molality? 58.44 grams of NaCl and you dissolved it in exactly 2.00 kg of pure water (the solvent). What... What is the molality when 0.75 mol is dissolved in 2.50 L of solvent? What is the molality when 48.0 mL of 6.00 M H2SO4 are diluted into 0.250 L? See all questions in Molality Impact of this question 3034 views around the world You can reuse this answer Creative Commons License