Question

How many grams of methyl alcohol should be added to $10$ litre tank of water to prevent its at $268K$?
($K_f$ for water is $1.84K$ $kg{mol}^{-1}$)

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

Answer 1

Answer: (C)

$868.06g$

By applying formula,

$△T_f=K_{f}m$

where m=molality

$273-268=1.86\times \frac{w}{M\times v}$

$5=1.86\times \frac{w}{32\times 10}w$

$=\frac{5\times 32\times 10}{1.86}=860.2\approx 868.06g$