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  • Boiling point of a aqueous solution of a non-volatile solute is equal to the boiling

Boiling point of a 2 \% aqueous solution of a non-volatile solute \mathrm{A} is equal to the boiling point of 8 \% aqueous solution of a non-volatile solute \mathrm{B}. The relation between molecular weights of \mathrm{A \: and \: B} is
 

Option: 1

\mathrm{M}_{\mathrm{A}}=4 \mathrm{M}_{\mathrm{B}}


Option: 2

\mathrm{M}_{\mathrm{B}}=4 \mathrm{M}_{\mathrm{A}}


Option: 3

\mathrm{M}_{\mathrm{A}}=8 \mathrm{M}_{\mathrm{B}}
 


Option: 4

\mathrm{M_{B}=8 M_{A}}


Answers (1)

best_answer

We know that
\mathrm{\Delta T_{b}= K_{b}m , m\: is\: molality. }

\mathrm{\Delta T_{b} } will be same for \mathrm{A} and \mathrm{B} due to solvent is same (water \rightarrow aqeous) and boilling point of aqueous solution of both \mathrm{A} and \mathrm{B} are equal.

\mathrm{K_{b}} will be due to solvent is same for \mathrm{A} and \mathrm{B}.
Hence,         \mathrm{\left ( \Delta T_{b} \right )_{A}= \left ( \Delta T_{b} \right )_{B}}
                     \mathrm{\left ( K_{b}m \right )_{A}= \left ( K_{b}m \right )_{B}}
                          \mathrm{m_{A}= m_{B}}
\mathrm{\frac{w_{A}}{\frac{M_{A}}{W_{solvent}}}= \frac{w_{B}}{\frac{M_{B}}{W_{solvent}}}}

Now , 2% aqueous solution of a non-volatile solute \mathrm{A} means in 100 ml or 100 g water 2 g of \mathrm{A} is present and 8% aqueous solution means in 100 ml or 100 g of water 8 g of \mathrm{B} is present.

So,
       \mathrm{\frac{2}{\frac{M_{A}}{100g}}= \frac{8}{\frac{M_{B}}{100g}}}

\mathrm{\Rightarrow M_{B}= 4\, M_{A}}

So, option (2) is correct.

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mansi

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