Question

45g of ethylene glycol is mixed with 600g of water. Calculate a. The freezing point depression and B. The freezing point of the solution

Answer 1

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Answer 2

Answer : (a) The freezing point depression is, 2.25 K

(b) The freezing point of the solution is, 270.5 K

Explanation :

(a) Formula used for lowering in freezing point :

$\Delta T_f= k_f\times m\\\\\Delta T_f= k_f\times \frac{w_2\times 1000}{M\times w_1}$

where,

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

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

m = molality

$w_1$ = mass of solvent (water) = 600 g

$w_2$ = mass of solute (ethylene glycol) = 45 g

M = molar mass of solute (ethylene glycol) = 62 g/mole

Now put all the given values in the above formula, we get the freezing point depression.

$\Delta T_f= (1.86K/moleKg)\times \frac{45g\times 1000}{62g/mole\times 600g}=2.25K$

(b) Now we have to calculate the freezing point of the solution.

$\Delta T_f=T_f^o-T_f\\\\T_f=T_f^o-\Delta T_f=273-2.25=270.5K$

Therefore, the freezing point of the solution is, 270.5 K