Question #2177b Chemistry Solutions Molality 1 Answer anor277 Jul 18, 2017 #chi_(NaOH)~=0.05# Explanation: Now.....#"Molality"# #=# #"Moles of solute"/"Kilograms of solvent."# And our #NaOH(aq)# solution is #3.0*"molal"#, i.e. #3.0*mol*kg^-1# Now #chi_(NaOH)="Moles of NaOH"/"Moles of NaOH + moles of water"# #=(3.0*mol)/(3.0*mol+(1000*g)/(18.01*g*mol^-1))# #=(3.0*mol)/(3.0*mol+55.52*mol)=0.0513# 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 1776 views around the world You can reuse this answer Creative Commons License