Question

75 g ethylene glycol (non-electrolyte) is dissolved in 500g water. The solution is placed in refrigerator maintained at a temperature of 263.7 k . What amount of ice is separate out at this temperature?

HINT:

[Kf water = 1.86 k molality^-1, i=1]​

Answer 1

Answer : The amount of ice separate out at this temperature is, 258.34 g

Explanation :

Formula used for lowering in freezing point :

$\Delta T_f=k_f\times m$

or,

$T^o_f-T_f=\frac{1000\times k_f\times w_2}{w_1\times M_2}$

where,

$T_f$ = freezing point of solution = 263.7 K

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

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

m = molality

$w_2$ = mass of solute (glucose) = 75 g

$w_1$ = mass of solvent (water) = ?

$M_2$ = molar mass of solute (ethylene glycol) = 62.07 g/mole

Now put all the given values in the above formula, we get the weight of solvent.

$(273-263.7)K=\frac{1000\times 1.86K/mole\times 75g}{w_1\times 62.07g/mole}$

$w_1=241.66g$

The amount of ice is separate out at this temperature = 500 - 241.66 = 258.34 g

Therefore, the amount of ice separate out at this temperature is, 258.34 g

Answer 2

Answer:

The amount of ice separate out at this temperature is, 258.34 g

Explanation :

Formula used for lowering in freezing point :

or,

where,

= freezing point of solution = 263.7 K

= freezing point of pure water = 273 K

= freezing point constant  = 1.86 K/mole

m = molality

= mass of solute (glucose) = 75 g

= mass of solvent (water) = ?

= molar mass of solute (ethylene glycol) = 62.07 g/mole

Now put all the given values in the above formula, we get the weight of solvent.

The amount of ice is separate out at this temperature = 500 - 241.66 = 258.34 g

Therefore, the amount of ice separate out at this temperature is, 258.34 g