Question

What is the osmotic pressure of a .333M solution of C9H12O6 at 25 Celsius?
What is the freezing point of a 1.77m solution of CBr4 in C6H6?
What is the boiling point of a 2.50m solution of C6H4Cl2 (non volatile)in CCl4? 
A solution of C2H2O4(non volatile) in C2H5OH has a boiling point of 80.38C. What is the morality of solution?

Accepted Answer

A) USING OSMOTIC PRESSURE FORMULA ,N=INRT/V​=IMRTFOR
MOLARITY M=1000+M×MW1000M×D​PUTTING THE
VALUES =(1000×1.25×342)
(1000×1.25×1.34)​=1.733MHENCE,N=IMRTN=28.6ATM    (FOR:I=1)

B) THE OSMOTIC PRESSURE OF SAME SOLUTION IS 0.474 ATM
ΔTF​=5.5−5.4=0.1OC
Π=CM​ST(C+M=MOLARITY)
ΔTF=CM×KF(CM=MOLALITY)FOR DILUTE SOLUTIONS CM​=CM​
∴Π=KF​
ΔTF​×ST=4.9
0.1×0.0821×283=0.474ATM

Question #146163

Expert's answer

A) Using osmotic pressure formula ,n=inRT/V=iMRTFor molarity M=1000+m×Mw1000m×dputting the values =(1000×1.25×342)
(1000×1.25×1.34)=1.733MHence,n=iMRTn=28.6atm (For:i=1)

B) The osmotic pressure of same solution is 0.474 atm
ΔTf=5.5−5.4=0.1oC
π=CMST(C+M=molarity)
ΔTf=Cm×Kf(Cm=molality)for dilute solutions CM=Cm
∴π=Kf
ΔTf×ST=4.9
0.1×0.0821×283=0.474atm

Question #146163

Expert's answer

A) Using osmotic pressure formula ,n=inRT/V​=iMRTFor molarity M=1000+m×Mw1000m×d​putting the values =(1000×1.25×342)(1000×1.25×1.34)​=1.733MHence,n=iMRTn=28.6atm (For:i=1)

B) The osmotic pressure of same solution is 0.474 atmΔTf​=5.5−5.4=0.1oCπ=CM​ST(C+M=molarity)ΔTf=Cm×Kf(Cm=molality)for dilute solutions CM​=Cm​∴π=Kf​ΔTf​×ST=4.90.1×0.0821×283=0.474atm

A) Using osmotic pressure formula ,n=inRT/V=iMRTFor molarity M=1000+m×Mw1000m×dputting the values =(1000×1.25×342)
(1000×1.25×1.34)=1.733MHence,n=iMRTn=28.6atm (For:i=1)

B) The osmotic pressure of same solution is 0.474 atm
ΔTf=5.5−5.4=0.1oC
π=CMST(C+M=molarity)
ΔTf=Cm×Kf(Cm=molality)for dilute solutions CM=Cm
∴π=Kf
ΔTf×ST=4.9
0.1×0.0821×283=0.474atm