$6.0 g$ of urea (molecular weight = 60) was dissolved in 9.9 moles of water. If the vapour pressure of pure water is $P^{\circ}$ , the vapour pressure of solution is:

0.10 P^{\circ}

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1.10 P^{\circ}

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0.90 P^{\circ}

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0.99 P^{\circ}

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Solution

The correct option is D $0.99 P^{\circ}$
$\frac{P^{\circ} - P}{P^{\circ}} = x_{s o l u t e} o r \frac{P^{\circ} - P}{P^{\circ}} = \frac{n}{n + N}$
$\Rightarrow \frac{0.1}{0.1 + 9.9} \Rightarrow 0.01; P = 0.99 P^{\circ}$

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Vapour pressure of pure water at 298 K is 23.8 mm Hg. 50 g of urea (NH₂CONH₂) is dissolved in 850 g of water. Calculate the vapour pressure of water for this solution and its relative lowering.

6.0 g

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(Molar mass = 60 g / m o l)

was dissolved in

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the vapour pressure of solution is:

Q. the vapour pressure of pure water is 60 mm hg at some temperature find the vapour pressure of solution containing 30g urea dissolved in 72g of water?

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(M = 60 g m o l^{- 1})

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Q. Vapour pressure of pure water at

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of urea

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6.0g of urea (molecular weight = 60) was dissolved in 9.9 moles of water. If the vapour pressure of pure water is P∘, the vapour pressure of solution is:

The correct option is D 0.99P∘P∘−PP∘=xsoluteorP∘−PP∘=nn+N⇒0.10.1+9.9⇒0.01;P=0.99P∘

$6.0 g$ of urea (molecular weight = 60) was dissolved in 9.9 moles of water. If the vapour pressure of pure water is $P^{\circ}$ , the vapour pressure of solution is:

The correct option is D $0.99 P^{\circ}$
$\frac{P^{\circ} - P}{P^{\circ}} = x_{s o l u t e} o r \frac{P^{\circ} - P}{P^{\circ}} = \frac{n}{n + N}$
$\Rightarrow \frac{0.1}{0.1 + 9.9} \Rightarrow 0.01; P = 0.99 P^{\circ}$