Directions : In the following question more than one of the answers given may be correct. Select the correct answer and mark it according to the code:
To 10 mL of\[5\text{ }M\text{ }BaC{{l}_{2}}\]solution, 5 mL of 0.5 M\[{{K}_{2}}S{{O}_{4}}\]solution is added. Barium sulphate precipitates out, what will happen? (1) Freezing point will decrease (2) Freezing point will increase (3) Boiling point will increase (4) Boiling point will decreaseA) 1, 2 and 3 are correct
B) 1 and 2 are correct
C) 2 and 4 are correct
D) 1 and 3 are correct
Correct Answer: C
Solution :
\[BaC{{l}_{2}}B{{a}^{2}}+2C{{l}^{-}}\] \[1\times 5=5\] \[2\times 5=10\]Initial mmol \[\therefore \]Total concentration\[=\frac{15}{10}=1.5\] [\[\because \]\[V=10\text{ }mL,\]given] \[{{K}_{2}}S{{O}_{4}}2{{K}^{+}}+SO_{4}^{2-}\] \[2\times 5\times 0.5\] \[5\times 0.5\] \[Initial\text{ }mmol\] \[=5\] \[=2.5\] When the two solutions are mixed, \[B{{a}^{2+}}+SO_{4}^{2-}\xrightarrow[{}]{{}}BaS{{O}_{4}}\] \[5\text{ }mmol\] \[2.5\text{ }mmol\] \[(5-2.5)\] \[=2.5\text{ }mmol\] Thus, the mmol of ions remaining in the solution \[=5({{K}^{+}})+2.5(B{{a}^{2+}})+10(C{{l}^{-}})=17.5\,mmol\] \[\therefore \]Total concentration\[=\frac{17.5}{15}=1.167\] Since, the number of solute particles decreases, the freezing point is raised and boiling point is lowered.
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