When 174.5 mg of octatomic sulfur (S_8) is a | vedclass

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(A) Given: Mass of solute $(w)$ = $174.5 \ mg = 0.1745 \ g$,Mass of solvent $(W)$ = $78 \ g$,Molar mass of $S_8$ $(m)$ = $8 	imes 32 = 256 \ g \ mol^

Vedclass · Accepted Answer

$332.19$

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(A) Given: Mass of solute $(w)$ = $174.5 \ mg = 0.1745 \ g$,Mass of solvent $(W)$ = $78 \ g$,Molar mass of $S_8$ $(m)$ = $8 	imes 32 = 256 \ g \ mol^{-1}$,$K_b = 5.2 \ K \ kg \ mol^{-1}$.The formula for elevation in boiling point is $\Delta T_b = \frac{1000 	imes K_b 	imes w}{m 	imes W}$.Substituting the values: $\Delta T_b = \frac{1000 	imes 5.2 	imes 0.1745}{256 	imes 78}$.$\Delta T_b = \frac{907.4}{19968} \approx 0.0454 \ K$.The boiling point of the solution = Boiling point of pure solvent + $\Delta T_b$.Boiling point = $332.15 + 0.0454 = 332.1954 \ K \approx 332.19 \ K$.

When $174.5 \ mg$ of octatomic sulfur $(S_8)$ is added to $78 \ g$ of bromine $(Br_2)$,the boiling point of bromine becomes ............. $K$. Given: $K_b$ for $Br_2 = 5.2 \ K \ kg \ mol^{-1}$ and the boiling point of pure $Br_2 = 332.15 \ K$.

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