How many grams of methyl alcohol should be added to 10L tank of water to prevent its freezing at 268 K? a)880.07 g b)899.04 g c)886.02 g d)868.06 g

Options

A. 880.07g
B. 899.04g
C. 886.02g
D. 868.06g

Answer

C

Explanation

Normal freezing point of water =273.15K. In order to prevent freezing at 268K, let the amount of methanol added be x g.
`therefore` Molality, `m= (x)/(32 xx 10)= (x)/(320)` [`because` Molar mass of `CH_(3)OH= 32g mol^(-1)` and volume of `H_(2)O=10L`]
Now, `Delta T_(f)= K_(f).m`
`(273.15-268) = 1.86 xx (x)/(320)`
`5.15 = 1.86 xx (x)/(320)` [`because K_(f)` for water =1.86 K kg `mol^(-1)`]
or `x= (5.15 xx 320)/(1.86)=886.02g`