Question

what mass of ethylene glycol ( molar mass 62.0 g/mol) must be added to 5.50 kg of water to lower the freezing point of water from 0° C to -100° C ??(Kf of water = 1.86 K kg/mol)​

Answer 1

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≈ 51,883.3 Kg

step-by-step explanation:

Given that,

Let,

M2 ( ethylene glycol) = 62.0 g/mol

and,

W1 (water) = 5.50 Kg

= 5500 g

Now,

∆Tf = 0° C - (-100°C) = 100 ° C = 283 K

Now,

W2 ( ethylene glycol ) = ??.

Given,

Kf = 1.86 K kg/mol

Now,

we know that,

∆Tf = (K1× W2 × 1000)/ (M2 × W1)

=> W2 = (∆Tf × M2 × W1)/(Kf × 1000)

=> W2 = (283 × 62 × 5500)/(1.86 × 1000)

=> W2 = 51,883.3 Kg

Answer 2

step-by-step explanation:

Given that,

Let,

M2 ( ethylene glycol) = 62.0 g/mol

and,

W1 (water) = 5.50 Kg

= 5500 g

Now,

∆Tf = 0° C - (-100°C) = 100 ° C = 283 K

Now,

W2 ( ethylene glycol ) = ??.

Given,

Kf = 1.86 K kg/mol

Now,

we know that,

∆Tf = (K1× W2 × 1000)/ (M2 × W1)

=> W2 = (∆Tf × M2 × W1)/(Kf × 1000)

=> W2 = (283 × 62 × 5500)/(1.86 × 1000)

=> W2 = 51,883.3 Kg