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 ?

Question

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

DrBob222 · Answer

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

lulu · Answer

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