`3.24 g` of sulphur dissolved in `40g` benzene, boiling point of the solution was higher than that of benzene by `0.081 K`. `K_(b)` for benzene is `2.53 K kg mol^(-1)`. If molecular formula of sulphur is `S_(n)`. Then find the value of `n`. (at.wt.of `S =32`).

On dissolving 3.24 g of sulphur in 40 g of benezene by 0.81 K. K_(b) value of benene is 2.53 k kg mol ^(-1) . Atomic mass of sulphur is 32 g mol ^(-1) . The molecular formula of sulphur is ______.

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) .

The molal freezing point depression constant for benzene is 4.9 K kg mol^(-1) . Selenium exists as a polymer of the type Se_(n) . When 3.26 g selenium is dissolved in 226 g of benzene, the observed freezing. If molecular formula of sulphur is S_(n) . Then find the value of n . (at. wt. of S = 32 ).

1.0 g of non-electrolyte solute dissolved in 50.0 g of benzene lowered the freezing point of benzene by 0.40 K . The freezing point depression constant of benzene is 5.12 kg mol^(-1) . Find the molecular mass of the solute.