The elevation in boiling point would be highest for:

0.08 M B a C l_{2}

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0.15 M K C l

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0.10 M

Glucose

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0.06 M C a (N O_{3})_{2}

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Solution

The correct option is C $0.15 M K C l$
$∵ Δ T_{b} = \frac{1000 \cdot K_{b} \cdot w}{m \cdot W} \times i$

$∴ Δ T_{b} \propto i \cdot \frac{w}{m \cdot \frac{W}{1000}} (∵ K_{b} = c o n s t a n t)$

or $Δ T_{b} \propto i \cdot M$

$(∵$ Molality $= \frac{w}{m \cdot \frac{W}{1000}}$ and assuming,

Molarity, M $=$ molarity $\approx$ m $=$ molality

Now, for the given solutions,

(a) For $0.08 M$ $B a C l_{2}$ , i $\cdot$ M $= 3 \times 0.08 = 0.24$

(b) For $0.15 M$ $K C l,$ i $\cdot$ M $= 2 \times 0.15 = 0.30$

(c) For $0.10 M$ glucose, i $\cdot$ M $= 1 \times 0.10 = 0.10$

(d) For $0.06 M$ $C a (N O_{3})_{2},$ i $\cdot$ M $= 3 \times 0.06 = 0.18$

Hence, $Δ T_{b}$ is maximum for $0.15 M$ $K C l$ solution.

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