The vapour pressure of methanol at a certain temperature is $1$ atm by adding a small amount of ethyl acetate the vapour pressure of the solution is found to be $684 m m$ . The relative lowering of vapour pressure is:

0.1

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0.9

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76

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0.694

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Solution

The correct option is A $0.1$
$1$ atm $= 760 m m$ of $H g$ pressure.

Now, relative lowering $= \frac{P^{0} - P_{s}}{P^{0}} = \frac{(760 - 684 m m)}{760 m m} = \frac{76}{760} = 0.1$

Option A is correct.

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The vapour pressure of methanol at a certain temperature is 1 atm by adding a small amount of ethyl acetate the vapour pressure of the solution is found to be 684 mm. The relative lowering of vapour pressure is:

The correct option is A 0.11 atm =760 mm of Hg pressure.Now, relative lowering =P0−PsP0=(760−684mm)760mm=76760=0.1 Option A is correct.

The vapour pressure of methanol at a certain temperature is $1$ atm by adding a small amount of ethyl acetate the vapour pressure of the solution is found to be $684 m m$ . The relative lowering of vapour pressure is:

The correct option is A $0.1$
$1$ atm $= 760 m m$ of $H g$ pressure.

Now, relative lowering $= \frac{P^{0} - P_{s}}{P^{0}} = \frac{(760 - 684 m m)}{760 m m} = \frac{76}{760} = 0.1$

Option A is correct.