A solution of 0.640 g of azulene in 100.0 g of benzene is `80.23^(@)C`. The boilingpoint of benzeneis `80.10^(@)C`, and` K_(b)` is `2.53^(@)C`/molal What is the moleculer mass of azulene?
A
108
B
99
C
125
D
134
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The correct Answer is:
c
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DILUTE SOLUTION
NARENDRA AWASTHI|Exercise Level 1 (Q.62 To Q.91)|1 Videos
A compound has the empirical formula C_(10)H_(8)Fe . A solution of 0.26 g of the compound in 11.2 g of benzene ( C_(6)H_(6) )boils at 80.26^(@)C . The boiling point of benzene is 80.10^(@)C , the K_(b) is 2.53^(@)C /molal. What is the molecules formula of the compound?
The freezing point of a solution of 2.40 g of biphenyl( C_(12)H_(10) ) in 75.0 g of benzene ( C_(6)H_(6) ) is 4.40^(@)C . The normal freezing point of benzene is 5.50^(@)C . What is the molal freezing point constant (^(@)C//m) for benzene ?
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) .
A solution containing 0.1g of a non-volatile organic substance P (molecular mass 100) in 100g of benzene raises the boiling point of benzene by 0.2^(@)C , while a solution containing 0.1 g of another non-volatile substance Q in the same amount of benzene raises the boiling point of benzene by 0.4^(@)C . What is the ratio of molecular masses of P and Q ?
0.5 m solution of acetic acid (Mw=60) in benzene (Mw=78) boils at 80.80^(@)C . The normal boiling point of benzene is 80.10^(@)C . And Delta_(vap)H=30.775 kJ mol^(-1) . Calculate the percent of association of acetic acid in benzene.
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) .
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 :
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.
NARENDRA AWASTHI-DILUTE SOLUTION-Level 3 - Match The Column
