Home
Class 12
CHEMISTRY
A solution containing 3.3g of a substanc...

A solution containing 3.3g of a substance in 125g of benzne (b.pt = `80^(@)C`) boils at `80.66^(@)C`. If `K_(b)` for benzene is `3.28 "K kg mol"^(-1)` the molecular mass of the substance will be :

A

127.20 g/mole

B

131.20 g/mole

C

137.12 g/mole

D

142.72 g/mole

Text Solution

Verified by Experts

The correct Answer is:
B
Promotional Banner

Similar Questions

Explore conceptually related problems

A solution containing 28 g of phosphorus in 315 g CS_(2)(b.p. 46.3^(@)C ) boils at 47.98^(@)C . If K_(b) for CS_(2) is 2.34 K kg mol^(-1) . The formula of phosphorus is (at , mass of P = 31).

A solution containing 28 g of phosphorus in 315 g CS_(2)(b.p.46.3^(@)C boils at 47.98^(@)C . If k_(b) for CS_(2) is 2.34 K kg mol^(-1) . The formula of phosphorus is (at .massof P= 31 ).

A solution containing 8 g of substances in 100 g of diethyl ether bolis at 36.86^(@)C , whereas pure ether boils at 35.60^(@)C . Determine the molecular mass of the solute. (For ether K_(b)2.02 K g "mol"^(-1))

If solution containing 0.15 g of solute dissolved in 15 g of solvent boils at a temperature higher by 0.216^(@)C than that of pure solvent, the molecular mass of the substance is (K_(b)=2.16^(@)C)

Pure benzene boiled at 80^(@)C . The boiling point of a solution containing 1 g of substance dissolved in 83.4 g of benzene is 80.175^(@)C . If latent heat of vaporization of benzene is 90 cal per g , calculated the molecular weight of solute.

A solution of 2.5g of non-volatile solid in 100g benzene is boiled at 0.42^(@)C higher than the boiling point of pure benzene. Calculate the molecular mass of the substance. Molal elevation constant of benzene is 2.67 "K kg mol"^(-1) .

when 1.80 g of a nonvolatile solute is dissolved in 90 g of benzene, the boiling point is raised to 354.11K . If the boiling point of benzene is 353.23K and K_(b) for benzene is 2.53 KKg mol^(-1) , calculate the molecular mass of the solute. Strategy: From the boiling point of the solution, calculate the boiling point elevation, DeltaT_(b) , then solve the equation DeltaT_(b)=K_(b)m for the molality m . Molality equals moles of solute divided by kilograms of solvent (benzene). By substituting values for molality and kilograms C_(6)H_(6) , we can solve for moles of solute. The molar mass of solute equals mass of solute (1.80 g) divided by moles of solute. The molecular mass (in amu) has the same numerical value as molar mass in g mol^(-1) .