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

The boiling point of water at 735 torr is 99.07°C. The mass of NaCl added in 100g water to make its boiling point 100°C is?(k=0.51 k kg/mol)

Open in App
Solution

Step 1:

Given that, kb=0.51KKgmol1

ΔTf=10099.07=0.93oC

Step 2:

Using the relation, ΔTf=Kf×m

m=0.930.51=1.824

nNaClWH2O=1.824

Step 3:

nNaCl=0.5×1.824×100×103=0.0911

mass = 0.0911×58.5=5.33g

Hence, the total amount of NaCl to be added is calculated as 5.33 g.


flag
Suggest Corrections
thumbs-up
36
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Concentration of Solutions
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon