Question

Ethylene glycol is used as an antifreeze in a cold cliamate Mass of ethylene glycol which should be added to `4` kg for water to prevent it from freezing at `-6^(@)C` will be (`K_(f)` for water =`1.86 K kg mol^(-1)` and molar mass of ethylene glycol =`62 g mol^(-1)`)
A. `400 g`
B. `800 g`
C. `304.60 g`
D. `204.30 g`

Answer 1

Correct Answer - 2
We have
`DeltaT_(f)=iK_(f)m`
`=(iK_(f)W_("solute"))/(M_("solute") W_("solvent"))`
Thus, `W_("solute")=(DeltaT_(f)M_("solute")W_("solvent"))/(iK_(f))`
we are given
`i=1`
`K_(f)=1.86 K kg mol^(-1)`
`DeltaT_(f)=T_(f^(0)-T_(f))=(0^(@)C)-(-6^(@)C)=6^(@)C= 6 K`
`M_("solute")=62 g mol^(-1)`
`W_("solvent")=4 kg`
Substituting these results, we get
`W_("solute")=((6 K)(62 g mol^(-1))(4 kg))/((1)(1.86 K kg mol^(-1))`
`=800 g`