The boiling point of 0-001 M aqueous solutions of N aCl - Na 2 SO 4- K 3 P O 4 and CH 3 COOH should follows the orderA- CH 3 COOH - NaCl - Na 2 SO 4- K 3 P O 4B- NaCl - Na 2 SO 4- K 3 PO 4- CH 3 COOHC- CH 3 COOH - K 3 P O 4- NaCl - Na 2 SO 4D- CH 3 COOH - K 3 PO 4- Na 2 SO 4- NaCl

The correct option is A CH_3COOH lt;NaCl lt; Na_2SO_4 lt; K_3PO_4T_b ∝ i× m Boiling point of solution∝ van't Hoff factor × molality Since number of ions ...

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Byju's Answer

Standard XII

Chemistry

Boiling Point of Carboxylic Acids

The boiling p...

Question

The boiling point of 0.001 M aqueous solutions of $N a C l, N a_{2} S O_{4}, K_{3} P O_{4} a n d C H_{3} C O O H$ should follows the order

C H_{3} C O O H < N a C l < N a_{2} S O_{4} < K_{3} P O_{4}

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N a C l < N a_{2} S O_{4} < K_{3} P O_{4} < C H_{3} C O O H

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C H_{3} C O O H < K_{3} P O_{4} < N a_{2} S O_{4} < N a C l

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C H_{3} C O O H < K_{3} P O_{4} < N a C l < N a_{2} S O_{4}

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Solution

The correct option is A $C H_{3} C O O H < N a C l < N a_{2} S O_{4} < K_{3} P O_{4}$
$△ T_{b} \propto i \times m$
$B o i l i n g p o i n t o f s o l u t i o n \propto v a n^{'} t H o f f f a c t o r \times m o l a l i t y$
Since number of ions is highest in aqueous solution of $K_{3} P O_{4}$ , so it will have the highest boiling point followed by aqueous solution of $N a_{2} S O_{4}$ and $N a C l$ . $C H_{3} C O O H$ is a weak electrolyte.
So, boiling point order is
$C H_{3} C O O H < N a C l < N a_{2} S O_{4} < K_{3} P O_{4}$

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The boiling point of 0.001 M aqueous solutions of NaCl, Na2SO4, K3PO4 and CH3COOH should follows the order

The boiling point of 0.001 M aqueous solutions of $N a C l, N a_{2} S O_{4}, K_{3} P O_{4} a n d C H_{3} C O O H$ should follows the order