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If 5.17 g of an unknown substance X, which is a nonelectrolyte, lowers the freezing point of 200g of benzene ( k = 4.9 C-/m) by 0.84 C what is the molecular weight of X?

Sagot :

sebassandin

Answer:

[tex]M.M=150.8g/mol[/tex]

Explanation:

Hello there!

In this case, according to the given information, it turns out firstly possible for us to set up the equation for the freezing point depression as shown below:

[tex]\Delta T=-i*m*Kf[/tex]

Thus, given Kf, i (equal to 1 because it is nonelectrolyte) and the freezing point depression, we can now calculate the molality of the solution:

[tex]m=\frac{\Delta T}{-i*Kf}=\frac{-0.84\°C}{-1*4.9\°C/m}\\\\m=0.17mol/kg[/tex]

Next, we calculate the kilograms of solvent by dividing the 200 g by 1000 to get 0.200 kg and thus calculate the moles of the solute X:

[tex]n_X=0.200kg*0.171mol/kg\\\\n_X=0.0343mol[/tex]

Finally, the molar mass by dividing the grams by moles:

[tex]M.M=\frac{5.17g}{0.0343mol}\\\\M.M=150.8g/mol[/tex]

Regards!

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