Question

How many gram of methyl alcohol should be added to 10 litre tank of water to prevent its freezing at 268k kf for water is 1.86?

Answer 1

Answer : The mass of methyl alcohol will be, 860.215 grams

Solution :

First we have to calculate the mass of solvent (water).

$\text{Mass of water}=\text{Density of water}\times \text{volume of water}=1kg/L\times 10L=10kg$

Formula used for lowering in freezing point :

$\Delta T_f= k_f\times m\\\\T^o_f-T_f= k_f\times \frac{w_2\times 1000}{M\times w_1}$

where,

$\DeltaT_f$ = change in freezing point  for freezing point depression

$T^o_f$ = freezing point of pure solvent  = 273 K

$T_f$ = freezing point of solution  = 268 K

$k_f$ = freezing point constant  = 1.86 K/m

$w_2$ = mass of solute (methyl alcohol) = ?

$w_1$ = mass of solvent (water) = 10 kg

M = molar mass of solute (methyl alcohol) = 32 g/mole

m = molality

Now put all the given values in this formula, we get the mass of methyl alcohol.

$273-268=1.86\times \frac{w_2}{32\times 10}$

$w_2=860.215g$

Therefore, the mass of methyl alcohol will be, 860.215 grams