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

A sample of industrial waste water is found to contain 8.2% Na3PO4 and 12% MgSO4 by weight in solution. If % ionisation of Na3PO4 and MgSO4 are 50% and 60% respectively, then its normal boiling point (in °C ) is (2 decimal places).
[use Kb(H2O) = 0.50 Kkgmo11]:


Open in App
Solution

For Na3PO4, i=1+3α =1+3×0.5=2.5For MgSO4, i=1+α =1+0.6=1.6
Let, 100g of water contain 8.2g Na3PO4 and 12g MgSO4ΔTb= Kbmi =Kb[effective no. of moles of (Na3PO4+MgSO4)wt. of solvent (in gm)×1000] ΔTb=0.50[8.2164×2.5+12120×1.679.8×1000]=1.785
Tb=100+1.785=101.785°C

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