1.

1.1g of CoCl_(3).6NH_(3) (mol.wt. =267) was dissolved in 100g of H_(2)O. The freezing point of the solution was -0.29^(@)C. How many moles of solute particles exist in solution for each mole of solute introduced ? K_(f) for H_(2)O=1.86^(@)C.m^(-1)

Answer»

Solution :Molality (experimental) `=(DeltaT_(f))/(K_(f))=(0.29)/(1.86)`
`=0.156` mole/1000g
Molality (THEORETICAL) `=("moles of solute")/("wt. of solvent in G")xx1000`
`=(1.1)/(267)xx(1000)/(100)=0.0412` mole/1000g
Thus, number of moles of solute PARTICLES produced by 1 mole of solute
`=(0.156)/(0.0412)=4`...........(Eqn. 10)


Discussion

No Comment Found

Related InterviewSolutions