Calculate the freezing point of a solution containing 45 grams of KCl and 4950.0

Question

Calculate the freezing point of a solution containing 45 grams of KCl and 4950.0 grams of water. The molal-freezing-point-depression constant (Kf) for water is 1.86 °C/m. Calculate the freezing point of a solution containing 45 grams of KCl and 4950.0 grams of water. The molal-freezing-point-depression constant () for water is 1.86 °C/m.

a. -0.45 °C

b. +0.45 °C

c. -0.23 °C

d. 1.23 °C

e. +0.23 °C

The concentration of S2O82- remaining at 400 s is ________ M.

The concentration of S2O82- remaining at 400 s is ________ M.

-0.007 +0.015 +0.035 +0.045 +0.057 The peroxyäsulfate ion (S208) reacts with the iodide ion in aqueous solution via the reaction: S2082-(aq) + 3r 2SO4 (aq) + 13" (aq) An aueous solution containing 0.050 M of520s2 ion and 0.072 MofT- is prepared, and the progress of the reaction followed by measuring [I-1 The data obtained is given in the table below. Time (s) 0.000 400.0 800.0 1200.0 1600.0 0.072 0.057 0.046 0.037 0.029

Explanation / Answer

Ans 1 : a) -0.45oC

Number of moles of KCl = 45 / molar mass

= 45 / 74.5513

= 0.604 mol

Molality = no. of moles of KCl / mass of solvent in kg

m = 0.604 / 4.950

= 0.122 m

Depression in freezing point = i. kf . m

i for KCl = 2

putting all the values we get :

= 2 x 1.86 x 0.122

= 0.45oC

So the freexing point of the solution will e 0 - 0.45

= -0.45oC

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