Elevation in boiling point was $0.52 \, ^\circ C$ when $6 \, g$ of a compound $X$ was dissolved in $100 \, g$ of water. The molecular weight of $X$ is: ($K_b$ for water is $0.52 \, K \, kg \, mol^{-1}$)

$A$ solution containing $2.5 \times 10^{-3} \ kg$ of a solute dissolved in $75 \times 10^{-3} \ kg$ of water boils at $373.535 \ K$. The molar mass of the solute is $..... \ g \ mol^{-1}$. [nearest integer] (Given: $K_b(H_2O) = 0.52 \ K \ kg \ mol^{-1}$,boiling point of water $= 373.15 \ K$)

Pure benzene boils at $80\,^oC$. When $1\,g$ of a solute is dissolved in $83.4\,g$ of benzene,the boiling point of the solution becomes $80.175\,^oC$. If the latent heat of vaporization of benzene is $90\,cal/g$,calculate the molar mass of the solute.

If $10 \ g$ of a solute (molar mass $= 100 \ g/mol$) is dissolved in $100 \ g$ of water,then the molal elevation constant $K_b$ of the solvent will be equal to:

What is molal elevation? Explain.

Which of the following aqueous solutions will have a boiling point of $102.2^{\circ} C$? The molal elevation constant for water is $K_b = 0.512 \ K \ kg \ mol^{-1}$. (Note: The provided constant $2.2 \ K \ kg \ mol^{-1}$ in the prompt is incorrect for water; using standard $K_b = 0.512 \ K \ kg \ mol^{-1}$ for calculation).