Q. 193.6( 32 Votes )

# A solution containing 30 g of non-volatile solute exactly in 90 g of water has a vapour pressure of 2.8 kPa at 298 K. Further, 18 g of water is then added to the solution and the new vapour pressure becomes 2.9 kPa at 298 K. Calculate:

i. molar mass of the solute

ii. vapour pressure of water at 298 K.

Answer :

given: mass of solute = 30g

Let the molar mass of solute be x g and vapour pressure of pure water at 298k be P_{1}^{⁰}

Mass of water(solvent) = 90g

Molar mass of water = H_{2}O = 1 × 2 + 16 = 18g

Moles of water = mass of water/molar mass

⇒ n = 90/18 moles

⇒ n = 5moles

Molar fraction of solute,

⇒

⇒

Vapour pressure of solution(p_{1}) = 2.8kpa

Applying the formula:

⇒

⇒

⇒

⇒ -(1)

According to second condition when we add 18g of water to solution vapour pressure becomes 2.9kpa

Moles of water = mass/molar mass

⇒ n = 90 + 18/18

⇒ n = 6moles

Molar fraction of solute,

⇒

⇒

Applying the formula:

⇒

⇒

⇒

⇒ -(2)

Solving 1 and 2:

Dividing (2) by (1) we get

⇒

⇒

0.8631(30 + 6x) = 30 + 5x

25.9 + 5.18x = 30 + 5x

0.18x = 4.1

X = 22.78g

Substituting value of x in 1 we get

⇒

⇒

⇒

⇒

⇒

⇒

i. molar mass of the solute = 22.78g

ii. vapour pressure of water at 298 K = 3.537kpa

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