The vapour pressure of pure benzene $C_{6} H_{6}$ at $50^{\circ} C$ is 268 torr. How many moles on non-volatile solute per mole of benzene is required to prepare a solution of benzene having a vapour pressure of 167 torr at $50^{\circ} C$ ?

0.377

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0.605

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0.623

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0.395

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Solution

The correct option is A
0.377

$p = P_{B}^{o} x_{B}$ (B = Benzene acts as solvent)
$x_{B} = \frac{167}{268} = 0.623 x_{s o l u t e} = 1 - x_{B} = 0.377$

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The vapour pressure of pure benzene $C_{6} H_{6}$ at $50^{\circ} C$ is 268 torr. How many moles on non-volatile solute per mole of benzene is required to prepare a solution of benzene having a vapour pressure of 167 torr at $50^{\circ} C$ ?

Q. The vapour pressure of pure benzene at

50^{\circ} C

268 t o r r

. How many mols of non-volatile solute per mol of benzene is required to prepare a solution of benzene having a vapour pressure of

167 t o r r

50^{\circ} C

Q. The vapour pressure of pure benzene at

50^{o} C is 268 m m of H g

. How many moles of non-volatile solute per mole of benzene are required to prepare a solution of benzene having a vapour pressure

167 m m of H g

50^{o} C

Q. The vapour pressure of pure benzene at

50^{o} C

is 268 mm of Hg. How many moles of non-volatile sohite per mole of benzene are required to prepare a solution of benzene having a vapour pressure 167 mm of Hg at

50^{o} C

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The vapour pressure of pure benzene C6H6 at 50∘C is 268 torr. How many moles on non-volatile solute per mole of benzene is required to prepare a solution of benzene having a vapour pressure of 167 torr at 50∘C?

The correct option is A 0.377 p=PoBxB (B = Benzene acts as solvent) xB=167268=0.623xsolute=1−xB=0.377

The vapour pressure of pure benzene $C_{6} H_{6}$ at $50^{\circ} C$ is 268 torr. How many moles on non-volatile solute per mole of benzene is required to prepare a solution of benzene having a vapour pressure of 167 torr at $50^{\circ} C$ ?

The correct option is A
0.377

$p = P_{B}^{o} x_{B}$ (B = Benzene acts as solvent)
$x_{B} = \frac{167}{268} = 0.623 x_{s o l u t e} = 1 - x_{B} = 0.377$