Assuming the formation of an ideal solution,d | vedclass

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(B) $1$. Calculate the moles of each component: $n_{	ext{benzene}} = \frac{1560}{78} = 20 \ mol$ $n_{	ext{chlorobenzene}} = \frac{1125}{112.5} = 10

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$100$

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(B) $1$. Calculate the moles of each component: $n_{	ext{benzene}} = \frac{1560}{78} = 20 \ mol$ $n_{	ext{chlorobenzene}} = \frac{1125}{112.5} = 10 \ mol$ $2$. Calculate the mole fractions: $\chi_{	ext{benzene}} = \frac{20}{20+10} = \frac{2}{3}$ $\chi_{	ext{chlorobenzene}} = \frac{10}{20+10} = \frac{1}{3}$ $3$. The boiling point is the temperature where the total vapor pressure $P_{	ext{total}} = P_{	ext{benzene}}^0 \chi_{	ext{benzene}} + P_{	ext{chlorobenzene}}^0 \chi_{	ext{chlorobenzene}} = 1000 \ torr$. $4$. Testing $100^{\circ}C$ from the graph: $P_{	ext{benzene}}^0 = 1350 \ torr$ $P_{	ext{chlorobenzene}}^0 = 300 \ torr$ $P_{	ext{total}} = (1350 	imes \frac{2}{3}) + (300 	imes \frac{1}{3}) = 900 + 100 = 1000 \ torr$. $5$. Since the total pressure equals the external pressure at $100^{\circ}C$,the boiling point is $100^{\circ}C$.

Assuming the formation of an ideal solution,determine the boiling point of a mixture containing $1560 \ g$ benzene (molar mass $= 78 \ g/mol$) and $1125 \ g$ chlorobenzene (molar mass $= 112.5 \ g/mol$) against an external pressure of $1000 \ torr$ using the provided graph. (in $^{\circ}C$)

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