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faust18 [17]
3 years ago
9

The normal freezing point of a certain liquid

Chemistry
1 answer:
stepladder [879]3 years ago
4 0

Answer : The molal freezing point depression constant of X is 4.12^oC/m

Explanation :  Given,

Mass of urea (solute) = 5.90 g

Mass of X liquid (solvent) = 450.0 g

Molar mass of urea = 60 g/mole

Formula used :  

\Delta T_f=i\times K_f\times m\\\\T^o-T_s=i\times K_f\times\frac{\text{Mass of urea}\times 1000}{\text{Molar mass of urea}\times \text{Mass of X liquid}}

where,

\Delta T_f = change in freezing point

\Delta T_s = freezing point of solution = -0.5^oC

\Delta T^o = freezing point of liquid X= 0.4^oC

i = Van't Hoff factor = 1  (for non-electrolyte)

K_f = molal freezing point depression constant of X = ?

m = molality

Now put all the given values in this formula, we get

[0.4-(-0.5)]^oC=1\times k_f\times \frac{5.90g\times 1000}{60g/mol\times 450.0g}

k_f=4.12^oC/m

Therefore, the molal freezing point depression constant of X is 4.12^oC/m

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F = (Kq1q2)/(r^2) where K = (9 × (10^9) Nm(C^-2))

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