Home
Class 12
CHEMISTRY
A solution containing 25.6 g of sulphur,...

A solution containing `25.6 g` of sulphur, dissolved in `1000 g` of naphthalene whose melting point is `80.1^(@)C` gave a freezing point lowering of `0.680^(@)C`. Calculate the formula of sulphur (`K_(f)` for napthalene =`6.8 K m^(-1)`)

Text Solution

Verified by Experts

`Delta_(f)=iK_(f)m`
`0.68= ixx 6.8 xx(25.6 xx1000)/(32 xx 1000)`
`i=1//8`
Sinve `i` is less than `1`, therefore, `S` is associated, i.e, 8 moles of `S` are associated as shown below
`8ShArrS_8`
Therefore, the molecular formula of sulphur is `S_(8)`.
Promotional Banner

Similar Questions

Explore conceptually related problems

Phenol associates in benzene to a certain extent to form a dimer. A solution containing 20 g of phenol in 1.0 kg of benzene has its freezing point lowered by 0.69 K. Calculate the fraction of phenol that has dimerised. [Given : K_(f) for benzene = 5.1 Km^(-1)]

1.355 g of a substance dissolved in 55 g of CH_(3)COOH produced a depression in the freezing point of 0.618^(@)C . Calculate the molecular weight of the substance (K_(f)=3.85)

By dissolving 13.6 g of a substance in 20 g of water, the freezing point decreased by 3.7^(@)C . Calculate the molecular mass of the substance. (Molal depression constant for water = 1.863K kg mol^(-1))

A solution containing 12.5 g of non-electrolyte solution in 175g of water gave a boiling point elevation of 0.7 k. calculate the molar mass of the solute if K_(b) for water is 0.52 K kg mol^(-1) .

When 0.6 g of urea is dissolved in 100 g of water, the water will boil at ( K_(b) for water = 0.52 Km^(-1) and normal boiling point of water = 100^(@)C ) :

The freezing point of a 0.08 molal solution of NaHSO^(4) is -0.372^(@)C . Calculate the dissociation constant for the reaction. K_(f) for water = 1.86 K m^(-1)

On dissolving 0.25 g of a non-volatile substance in 30 mL benzene (density 0.8 g mL^(-1)) , its freezing point decreases by 0.25^(@)C . Calculate the molecular mass of non-volatile substance (K_(f) = 5.1 K kg mol^(-1)) .

The molal freezing point depression constant of benzene (C_(6)H_(6)) is 4.90 K kg mol^(-1) . Selenium exists as a polymer of the type Se_(x) . When 3.26 g of selenium is dissolved in 226 g of benzene, the observed freezing point is 0.112^(@)C lower than pure benzene. Deduce the molecular formula of selenium. (Atomic mass of Se=78.8 g mol^(-1) )

The freezing point of a solution contaning 0.3 g of acetic acid in 43 g of benzene reduces by 0.3^(@) . Calculate the Van's Hoff factor "( K_(f) for benzene = 5.12 K kg mol^(-1) )"

Addition of 0.643 g of compound to 50 ml of liquid (density 0.879 g/ml) lowers the freezing point from 5.51^@C to 5.03^@C . Calculate the molar mass of the compound. [ K_f for benzene =5.12 K Kg mol^(-1) ]