Question

If
`m=` molality of solution `(mol Kg^(-1))`
`chi_(solute)=` mole fraction of solute
`M_(solvent)=` molar mass of solvent `(g mol^(-1))`
A. `chi_(solute)=1000+ mM_(solvent)//mM_(solvent)`
B. `chi_(solute)=mM_(solvent)//mM_(solvent)-1000`
C. `chi_(solute)=(mM_(solvent))/(1000+mM_(solvent))`
D. `chi_(solute)=1000 mM_(solvent)//mM_(solvent)`

Answer 1

Correct Answer - 3
`chi_(solute)=n_(solute)/(n_(solute)+n_(solvent))`
`m=n_(solute)/(Kg_(solvent))`
Since solution is a homogeneous mixture, we can consider any size of solution. If we take `1 kg` of solvent, then
`m=n_(solute)`
and
`n_(solvent)=(1000 g)/(M_(solvent))`
where `M_(solvent)` is the molar mass of solvent. Substuting this result into fraction expression, we get
`chi_(solute)=m/(m+1000/M_(solv))`
`=(mM_(solv))/(mM_(solv)+1000)`