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Grade: 11
        
Derive relation between molality and mole fraction
 
3 years ago

Answers : (3)

nishchith
30 Points
							

XA = n/N+n and XB = N/N+n    

 XA/XB = n/N = Moles of solute/Moles of solvent = wA/mB/wB×mA    

XA×1000/XB×mB = wA×1000/wB×mA = m    

or

\small \frac{X_A\times 1000}{1-X_A} = m

3 years ago
Akanksha Hansika
23 Points
							
You went somewhere wrong. The correct relation is
(X1000) / [(1-Xax mB] = m
3 years ago
Ravi Kumar Singh
11 Points
							Xa= na/na+nb,Xb=nb/na+nbXa/Xb=na/nb=Wa*Ma/Wb*MaXa*1000/Xb*Mb=Wa*1000/Wb*Ma=na*1000/Wb=mXa*1000/(1-Xa)*mb=m
						
6 months ago
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