Home
Class 12
CHEMISTRY
0.90g of a non-electolyte was dissolved ...

0.90g of a non-electolyte was dissolved in 87.9 g of benzene. This reised the boiling point of benzene by `0.25^(@)C`. If the molecular mass of the non-electrolyte is 103.0 g `mol^(-1)` , calculate the molal elecation constant for benzene.

Text Solution

AI Generated Solution

To calculate the molal elevation constant (Kb) for benzene, we can follow these steps: ### Step 1: Gather the given data - Mass of the non-electrolyte (solute), \( W_b = 0.90 \, \text{g} \) - Mass of benzene (solvent), \( W_a = 87.9 \, \text{g} \) - Elevation in boiling point, \( \Delta T_b = 0.25 \, ^\circ C \) (which is equivalent to 0.25 K) - Molecular mass of the non-electrolyte, \( M_b = 103.0 \, \text{g/mol} \) ...
Promotional Banner

Similar Questions

Explore conceptually related problems

0.75 g of a non-electrolyte was dissolved in 87.9 g of benzene. This raised the boiling point of benzene by 0.25^(@)C . If the molecular mass of non-electrolyte is 103 , calculate the molal elevation constant for bezene.

0.440 g of a substance dissolved in 22.2 g of benzene lowered the freezing point of benzene by 0.567 ^(@)C . The molecular mass of the substance is _____ (K_(f) = 5.12 K kg mol ^(-1))

10 gram of a non -volatile solute when dissolved in 100 gram of benzene raises its boiling point 1^(@) . What is the molecular mass of the solute ? ( k_(b) for benzene K mol^(-1) ).