Question

A 5% solution of cane sugar (Mol. wt. = 342) is isotonic with 1% solution of X under similar conditions. The mol. wt. of X is

(a) 136.2 (b) 68.4 (c) 34.2 (d) 171.2​

Answer 1

$\large \rm {\underbrace{\underline{Elucidation:-}}}$

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$\sf \red {\underline{\underline{Provided\: that:}}}$

➻Two solutions are isotonic to each other .

➻Solution of cane sugar=5%

➻Molecular weight of cane sugar=342

➻Solution of x=1%

➻Solution of cane sugar is isotonic with solution of x

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$\sf \blue {\underline{\underline{To\: determine:}}}$

➻Molecular weight of x=?

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$\sf \pink {\underline{\underline{We\: Know:}}}$

$\sf {\boxed{Molality(m)=\frac{No\: of\: moles\: of\: solute}{Volume\: of\: solvent\: in\: kg}}}$

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★Solution 1:-

➻Solution of cane sugar=5%

➻That means 5g of solute is present in 100 grams of water.

➻Weight of solute in grams=5

➻Molecular weight=342

$\large \to \sf {No\: of\: moles=\frac{Weight\: in\: grams}{Molecular\: weight}}$

$\large \to \sf {No\: of\: moles=\frac{5}{342}}$

$\to \sf {No\: of\: moles=0.014}$

➻Volume of water =100grams

➻As we want the water in kg,

$\sf {\frac{100}{1000}}$

$\implies \sf {Volume\: of\: water=0.1kg}$

$\large \sf {Molality_{(Solution\:1)}=\frac{No\: of\: moles\: of\: solute}{Volume\: of\: solvent\: in\: kg}}$

$\large \to \sf {Molality_{(Solution\:1)}=\frac{0.0146}{0.1}}$

$\implies \sf \green {\boxed{\underline{Molality_{(Solution\:1)}=0.146}}}$

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★Solution 2:-

➻Solution of x=1%

➻That means 1g of solute is present in 100 grams of water.

➻Weight of solute in grams=1

➻Molecular weight=M(to be determined)

➻Volume of water =100grams

$\large \to \sf {No\: of\: moles=\frac{Weight\: in\: grams}{Molecular\: weight}}$

$\to \sf {No\: of\: moles=\frac{1}{M}}$

➻Volume of water =100grams

$\implies \sf {Volume\: of\: water=0.1kg}$

$\large \sf {Molality_{(Solution\:2)}=\frac{No\: of\: moles\: of\: solute}{Volume\: of\: solvent\: in\: kg}}$

$\large \to \sf {Molality_{(Solution\:2)}=\frac{\frac{1}{M}}{0.1kg}}$

$\implies \sf \green {\boxed{\underline{Molality_{(Solution\:2)}=\frac{1}{0.1M}}}}$

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$\sf \purple {\underline{\underline{According\: to\: given\: condition:}}}$

➻Solution of cane sugar is isotonic with solution of x

➻Solution 1 is isotonic with solution 2

➻Molarity of solution 1=Molality of solution 2

➻Equating the two Molalities of the solutions

$\to \sf {Molality_{(Solution\:1)}=Molality_{(Solution\:2)}}$

$\large \to \sf {0.146=\frac{1}{0.1M}}$

$\large \to \sf {M=\frac{1}{0.1×0.146}}$

$\large \to \sf {M=\frac{1×10×1000}{1×146}}$

$\large \to \sf {M=\frac{10000}{146}}$

$\to \sf {M≈68.493}$

$\implies \sf \green {\boxed{\underline{Molecular\: weight\: of\: x=68.4}}}$

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$\sf \orange {\underline{\underline{Therefore,}}}$

➻Molecular weight of x is 68.4

★"Option-B" is the correct option.

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