A) \[1.273\text{ }g/c{{m}^{3}}\]
B) \[2.173\text{ }g/c{{m}^{3}}\]
C) \[1.133\text{ }g/c{{m}^{3}}\]
D) \[2.612\text{ }g/c{{m}^{3}}\]
Correct Answer: C
Solution :
Volume of ethylene glycol \[=50\text{ }c{{m}^{3}}\] \[\therefore \]Mass of ethylene glycol \[(w)=50\times d\] And, \[W=50g\] \[\Delta T={{0}^{-}}(-34)=34\] \[{{K}_{f}}=1.86K\,\log mo{{l}^{-1}}\] \[\therefore \] \[\Delta {{T}_{f}}=\frac{1000\times {{k}_{f}}\times w}{62\times 50}\] (Mol. mass of ethylene glycol= 62) or \[d=\frac{34\times 62\times 50}{1000\times 1.86\times 50}\] or \[d=1.33\,g/c{{m}^{3}}\]
You need to login to perform this action.
You will be redirected in
3 sec
