Question

0.75mol of ethylene bromide were mixed with 0.25 mol of propylene bromide at 358K to form nearly ideal solution. Vapour pressure of pure ethlene bromide and propylene bromide at 358K are 2.77?10'4N m-2 and 1.73?10'4N m-2 respectively. Calculate the vapour pressure of the solution.

Answer 1

4 hari lalu · Vapour pressure of pure ethlene bromide and propylene bromide at 358K are 2.77?10'4N m-2 and .

Answer 2

The vapour pressure of the solution is $2.51\times 10^4Nm^{-2}$

Explanation:

According to Raoult's law, the vapor pressure of a component at a given temperature is equal to the mole fraction of that component multiplied by the vapor pressure of that component in the pure state.

$p_1=x_1p_1^0$ and $p_2=x_2p_2^0$

where, x = mole fraction and

$p^0$ = pressure in the pure state

$x_{\text {ethylene bromide}}=\frac{0.75}{0.75+0.25}=0.75$

$x_{\text {propylene bromide}}=\frac{0.25}{0.75+0.25}=0.25$

According to Dalton's law, the total pressure is the sum of individual pressures.

$p_{total}=p_1+p_2$

$p_{total}=x_{\text {ethylene bromide}}\times p_{\text {ethylene bromide}}^0+x_{\text {ethylene bromide}}\times p_{\text {ethylene bromide}}^0$

$p_{total}=0.75\times 2.77\times 10^4+0.25\times 1.73\times 10^4$

$p_{total}=2.51\times 10^4Nm^{-2}$

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