Post a New Question

Posted by lucy on Saturday, October 9, 2010 at 4:45am.

The boiling point of water at 735mmHg is 99.073°C. What mass of BaCl2 should be added to 2.28kg water to increase the boiling point to 100.000°C ? Kb for water = 0.510 K kg mol-1 2589 g 777 g 734 g 288 g 863 g

delta T = i*Kb*m Solve for m molality = moles/kg Solve for moles moles = grams/molar mass Solve for grams.

dT= 100- 99.073= 0.927 dT= m Kb i 0.972= m*0.51*3= 0.972/0.510*3 =0.606 mol BaCl2/ 2.28kg 0.606*2.28kg= 1.38g mass BaCl2= 1.38g*208.23g/mol =288g

More Related Questions