Question

An aqueous solution of 2% non-volatile Solute exerts a pressure of 1.004 bar at the normal boilir
point of the solvent. What is the Molar mass of solute--​

Answer 1

Answer:

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? Hence, the molar mass of the solute is 41.35 g mol - 1

Explanation:

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    =  (w2 x M1 ) / (M2  x 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.

Answer 2

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 = (w2 x M1 ) / (M2 x 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

Explanation:

mark as brainleast