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Question:
An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solvent. What is the molar mass of the solute?

Answer:

Here,

Vapour pressure of the solution at normal boiling point (p₁) = 1.004 bar (Given)

Vapour pressure of pure water at normal boiling point (p₁⁰) = 1.013 bar

Mass of solute, (w₂) = 2 g

Mass of solvent (water), (w₁) = 100 - 2 = 98 g

Molar mass of solvent (water), (M₁) = 18 g mol^{- 1}

According to Raoult's law,

(p₁⁰ - p₁) / p₁⁰= (w₂x M₁) / (M₂x w₁ )

(1.013 - 1.004) / 1.013 = (2 x 18) / (M₂ x 98 )

0.009 / 1.013 = (2 x 18) / (M₂ x 98 )

M₂ = (2 x 18 x 1.013) / (0.009 x 98)

M₂ = 41.35 g mol^{- 1}

Hence, the molar mass of the solute is 41.35 g mol^{- 1}.

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Question:
An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solvent. What is the molar mass of the solute?

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Question: An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solvent. What is the molar mass of the solute?

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Question:
An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solvent. What is the molar mass of the solute?