The variation of vapour pressure (b) as a fun | vedclass

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(C) The boiling point of a liquid is the temperature at which its vapour pressure becomes equal to the external atmospheric pressure $(760 \ mm \ Hg)$

Vedclass · Accepted Answer

$CCl_4, C_2H_5OC_2H_5, H_2O$

Vedclass · Answer

(C) The boiling point of a liquid is the temperature at which its vapour pressure becomes equal to the external atmospheric pressure $(760 \ mm \ Hg)$.Given boiling points:$C_2H_5OC_2H_5 = 308 \ K$$CCl_4 = 350 \ K$$H_2O = 373 \ K$From the graph,by drawing perpendiculars from the intersection points of the curves with the $760 \ mm \ Hg$ line to the temperature axis $(a)$,we observe the order of boiling temperatures as $T_B Substituting the given values:$T_B = 308 \ K$ (for $C_2H_5OC_2H_5$)$T_A = 350 \ K$ (for $CCl_4$)$T_C = 373 \ K$ (for $H_2O$)Therefore,curve $A$ corresponds to $CCl_4$,curve $B$ corresponds to $C_2H_5OC_2H_5$,and curve $C$ corresponds to $H_2O$.The correct sequence for $A, B, C$ is $CCl_4, C_2H_5OC_2H_5, H_2O$.

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