wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Calculate the boiling point of solutions when 2 g of Na2SO4 (M=142g mol1) is dissolved in 50 g of water, assuming Na2SO4 undergoes complete ionization. ( Kb for water = 0.52 K kg mol1 )

Open in App
Solution

Moles of Na2SO4=2g142 g mol1=0.014 mol

Number of kilograms of solvent =50g1000 g kg1

=0.02kg

Thus, molality of Na2SO4 solution =0.014 mol0.02 kg

=0.7 mol kg1

For water, change in boiling point,

Tb=Kb×m

Tb=0.52 K kg mol1 × 0.07 mol kg1

Tb=0.0364 K

Since water boils at 373.15 K at 1.013 bar pressure, the boiling point of solution will be 373.15 + 0.0364 = 373.1864 K.

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Osmotic Pressure
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon