Question

1.5 gm of a non-volatile, non-electrolyte is dissolved in 50 gm benzene (Kb = 2.5 k Kg/mole). The elevation of boiling point of the solution is 0.75 K. The molecular weight of the solute in gm/mole is -​

Answer 1

Answer:

$\bigstar{\bold{Molar\:mass=100\:gm/mol}}$

Explanation:

$\Large{\underline{\underline{\bf{Given:}}}}$

• Weight of solute (w₂) = 1.5 gm
• Weight of solvent (w₁) = 50 gm
• Elevation in boiling point ($\Delta\:T_b$) = 0.75 K
• Molal elevation constant ($K_b$) = 2.5 K kg/mol

$\Large{\underline{\underline{\bf{To\:Find:}}}}$

• Molar mass of the solute (M₂)

$\Large{\underline{\underline{\bf{Solution:}}}}$

→ Here we have to find the molar mass of the solute dissolved in benzene.

→ The relation between molar mass of the solute and the elevation in boiling point is given by the formula,

  $\sf {M_2=\dfrac{1000\times w_2\times K_b}{\Delta T_b\times w_1}}$

→ Substituting the given datas in the above formula, we get the molar mass of the solute.

   $\sf {M_2=\dfrac{1000\:g/kg\times 1.5\:gm\times 2.5\:K\:kg/mol}{0.75\:K\times 50\:gm}}$

→ Simplifying,

   $\sf{M_2=\dfrac{3750\:gm}{37.5\:mol} }$

  M₂ = 100 g/mol

→ Hence the molar mass of the solute is 100 gm/mol

$\boxed{\bold{Molar\:mass=100\:gm/mol}}$

$\Large{\underline{\underline{\bf{Notes:}}}}$

→ The molal elevation constant is also called as the boling point elevation constant or Ebullioscopic constant.

→ The unit of boiling point elevation constant is K kg/mol .

→ The boiling point of a solution is always higher than the boling point of the pure solvent.

→ The elevation of boilig point is directly proportional to the molal concentration of the solute present in the solution.

  $\Delta T_b\propto m$

  $\Delta T_b = K_b\:m$

 where $K_b$ is the molal elevation constant.