651 g of ethylene glycol (HOCH_2CH_2OH) is d | vedclass

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Question

(B) The molar mass of ethylene glycol $(HOCH_2CH_2OH)$ is $62 \ g/mol$. Number of moles of ethylene glycol $(n_{solute}) = \frac{651 \ g}{62 \ g/mol}

Vedclass · Accepted Answer

$0.62$

Vedclass · Answer

(B) The molar mass of ethylene glycol $(HOCH_2CH_2OH)$ is $62 \ g/mol$. Number of moles of ethylene glycol $(n_{solute}) = \frac{651 \ g}{62 \ g/mol} = 10.5 \ mol$. Number of moles of water $(n_{solvent}) = \frac{1500 \ g}{18 \ g/mol} \approx 83.33 \ mol$. Using Raoult's law for a non-volatile solute: $\frac{P^{\circ} - P}{P^{\circ}} = \frac{n_{solute}}{n_{solute} + n_{solvent}}$. $\frac{0.7 - P}{0.7} = \frac{10.5}{10.5 + 83.33} = \frac{10.5}{93.83} \approx 0.1119$. $0.7 - P = 0.7 	imes 0.1119 \approx 0.0783$. $P = 0.7 - 0.0783 = 0.6217 \ atm$. Thus,the vapour pressure is closest to $0.62 \ atm$.

$651 \ g$ of ethylene glycol $(HOCH_2CH_2OH)$ is dissolved in $1.5 \ kg$ of water at $363 \ K$. The vapour pressure of pure water at $363 \ K$ is $0.7 \ atm$. Assuming ideal solution behaviour,the vapour pressure of water over the solution (in $atm$) is closest to $...$

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