A solution containing 2.5 g of a non-volatile solute in 100 gm of benzene boiled at a temperature 0.42K higher than at the pure solvent boiled. What is the molecular weight of the solute? The molal elevation constant of benzene is "2.67 K kg mol"^(-1).

A solution containing 2.5 g of a non-volatile solute in 100 gm of benz

1.	A solution containing 2.5 g of a non-volatile solute in 100 gm of benzene boiled at a temperature 0.42K higher than at the pure solvent boiled. What is the molecular weight of the solute? The molal elevation constant of benzene is "2.67 K kg mol"^(-1).
Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :`K_(b)="<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>.67 K <a href="https://interviewquestions.tuteehub.com/tag/kg-1063886" style="font-weight:bold;" target="_blank" title="Click to know more about KG">KG</a> mol"^(-1)` <br/> `DeltaT_(b)=0.42K` <br/> `W_(1)=100g=(100)/(1000)kg=0.1Kg` <br/> `W_(2)=(K_(b))/(DeltaT_(b)).(W_(2))/(W_(1))` <br/> `M_(2)=(2.67)/(0.42)<a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a>(2.5)/(0.1)` <br/> `M_(2)="158.98 g mol"^(-1)`</body></html>