Question

5. Which of the following solutions would most likely have the highest boiling point?
(A) 0.100 M KOH
(B) 0.100 M Na2SO4
(C) 0.200 M CaCl2 (D) 0.100 M C6H12O6

Answer 1

Answer:

c

Explanation:

c

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answer

ok

?c

cccccc

Answer 2

The solution having highest boiling point is 0.1 m $CaCl_2$

Explanation:

The expression of elevation in boiling point is given as:

$\Delta T_b=i\times k_b\times m$

where,

$\Delta T_b$ = Elevation in boiling point

i = Van't Hoff factor  

$T_b$ = change in boiling point

$k_b$ = boiling point constant

m = molality

For the given options:

• Option A:  0.1 m KOH

Value of i = 2

So, molal concentration will be = $(0.1)\times 2=0.2m$

• Option B:  0.100 m $Na_2SO_4$

Value of i = 3

So, molal concentration will be = $(0.100)\times 3=0.300m$

• Option C:  0.200 m $CaCl_2$

Value of i = 3

So, molal concentration will be = $(0.200)\times 3=0.600m$

• Option D:  0.100 m $C_6H_{12}O_6$

Value of i = 1  (for non-electrolytes)

So, molal concentration will be = $(0.100)\times 1=0.100m$

As, the molal concentration of  is the highest, so its boiling point will be the highest.

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