How much `C_(2)H_(5)OH` must be added to `1.0″L of “H_(2)O`, so that solution should not freeze at `-4^(@)F` ?
`[K_(f)(C_(2)H_(5)OH)=1.86^(@)C//m]`
A. `lt10.75g`
B. `gt494.5g`
C. `lt20g`
D. `494.5g`
`[K_(f)(C_(2)H_(5)OH)=1.86^(@)C//m]`
A. `lt10.75g`
B. `gt494.5g`
C. `lt20g`
D. `494.5g`
Correct Answer – B
`(C)/(5)=(F-32)/(9)`
`-4^(@)F=-20^(@)CrArrDeltaT_(f)=20^(@)C`
`”From “DeltaT_(f)=K_(f).mrArrm=(20)/(1.86)=10.75m`
`”Mass of “C_(2)H_(5)OH” needed “=10.75xx46=494.5g`
Solution not to freeze at `-20^(@)C` will require more ethanol than the mass needed to freeze at `-20^(@)C`.