At T(K),the vapour pressures of two liquids,h | vedclass

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(A) $1$. Calculate the molar masses: Heptane $(C_7H_{16})$ = $7 	imes 12 + 16 	imes 1 = 100 \ g/mol$. Octane $(C_8H_{18})$ = $8 	imes 12 + 18 	ime

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$66.66$

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(A) $1$. Calculate the molar masses: Heptane $(C_7H_{16})$ = $7 	imes 12 + 16 	imes 1 = 100 \ g/mol$. Octane $(C_8H_{18})$ = $8 	imes 12 + 18 	imes 1 = 114 \ g/mol$. $2$. Calculate the number of moles: $n_{	ext{heptane}} = \frac{25 \ g}{100 \ g/mol} = 0.25 \ mol$. $n_{	ext{octane}} = \frac{57 \ g}{114 \ g/mol} = 0.50 \ mol$. $3$. Calculate the mole fractions: $x_{	ext{heptane}} = \frac{0.25}{0.25 + 0.50} = \frac{0.25}{0.75} = \frac{1}{3}$. $x_{	ext{octane}} = \frac{0.50}{0.75} = \frac{2}{3}$. $4$. Apply Raoult's Law: $P_{	ext{total}} = P^{\circ}_{	ext{heptane}} 	imes x_{	ext{heptane}} + P^{\circ}_{	ext{octane}} 	imes x_{	ext{octane}}$. $5$. $P_{	ext{total}} = 106 	imes \frac{1}{3} + 47 	imes \frac{2}{3} = \frac{106 + 94}{3} = \frac{200}{3} = 66.66 \ kPa$.

At $T(K)$,the vapour pressures of two liquids,heptane and octane are $106 \ kPa$ and $47 \ kPa$ respectively. If $25 \ g$ of heptane and $57 \ g$ of octane form an ideal solution,at $T(K)$ the vapour pressure of solution in $kPa$ is :

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