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45 g of ethyl...

Question

$45 g$ of ethylene glycol $C_{2} H_{6} O_{2}$ is mixed with $600 g$ of water. Calculate:

$A$ : The freezing point depression.

$B$ : The freezing point of the solution.

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Solution

$A$ :

Molality of solution
$M o l a l i t y = \frac{mass of ethylene glycol(g)}{molar mass of ethylene glycol \times mass of water(kg)}$

$M o l a l i t y = \frac{45 g}{62 g m o l^{- 1} \times 600 \times 10^{- 3} k g}$

$M o l a l i t y = 1.2 m o l k g^{- 1}$

Depression in freezing point

Depression in freezing point $(Δ T_{f}) = K_{f} \times m o l a l i t y (m)$

We know that,

Freezing point depression constant $(K_{f}) = 1.86 K k g m o l^{- 1}$

By putting the values, we get,
$Δ T_{f} = 1.86 K k g m o l^{- 1} \times 1.2 m o l k g^{- 1} = 2.2 K$

$B$ :

Molality of solution
Molality of solution
$M o l a l i t y = \frac{mass of ethylene glycol(g)}{molar mass of ethylene glycol \times mass of water(kg)}$

$M o l a l i t y = \frac{45 g}{62 g m o l^{- 1} \times 600 \times 10^{- 3} k g}$

$M o l a l i t y = 1.2 m o l k g^{- 1}$

Depression in freezing point

Depression in freezing point $(Δ T_{f}) = K_{f} \times molality(m)$

We know that,

Freezing point depression constant $(K_{f}) = 1.86 K k g m o l^{- 1}$

By putting the values, we get,
$Δ T_{f} = 1.86 K k g m o l^{- 1} \times 1.2 m o l k g^{- 1} = 2.2 K$

Freezing point of solution
We know that $Δ T_{f} = T_{f}^{0} - T_{f}$

We also know that freezing point of pure
water, $T_{f}^{0} = 0^{0} C or 273.15 K$

Freezing point of the aqueous solution $(T_{f}) = 273.15 K - 2.2 K = 270.95 K$

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Similar questions

45

(C_{2} H_{6} O_{2})

600

K_{f}

1.86

m o l^{- 1}

45 g

600 g

k_{f} = 1.86 K k g m o l^{- 1}

31 g

(C_{2} H_{6} O_{2})

500 g

K_{f}

2 K

k g

{m o l}^{- 1}

K

= 273 K

C_{2} H_{4} (O H)_{2}

1.113 g c m^{- 3}

(K_{f} o f H_{2} O = 1.86 K k g m o l^{- 1})

83 g

625 g

K

= 1.86 K k g m o l^{- 1} .

= 273 K,

= C : 12.0 u, O : 16.0 u, H : 10 u]

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