A quantity of `0.25g` of a substance when vaporized displaced `50cm^(3)` of air at `0^(0)C` and `1` atm.The gram molecular mass of the substance will be

0.1693 gm of a volatile substance when vapourised displaced 58.9 cm of air measured at 27^@C and 746 mm pressure. Calculate the molecular mass of the substance. (Aqueous tenstion at 27^@C=26.7mmHg .)

The boiling point of a solution of 0.11g of a substance is 15g of ether was found to be 0.1^(@)C higher than that of pure ether. The molecular weight of the substance will be (K_(b) = 2.16)

0.15g of substance displaced 58.9cm^(3) of air at 300K and 746mm pressure Calculate the molecular mass ("Aq Tension at" 300K = 26.7mm) .

M gram of a substance when vapourised occupies of 5.6 litre at NTP. The molar mass (in g/mol) of the substance will be :

The boiling point of a solution of 0.11 of a substance in 15g of ether was found to be 0.1^(@)C higher than that of pure ether. The molecular weight of the substance will be (K_(b)=2.16) :

(a) A solution containing 0.5g of naphithalene in 50g C Cl_(4) yield a boiling point elevation of 0.4K , while a solution of 0.6g of an unknown solute in the same mass of the solvent gives a boiling point elevation of 0.65K . Find the molar mass of the unknown solute. (b) The boiling point of a solution of 0.1g of a substance in 16g of ether was found to be 0.100^(@)C higher that of pure ether. What is the molecular mass of the substance. K_(b) (ether) =2.16K kg "mol"^(-1)

In Victor Meyers method, 0.2 g of an organic substance displaced 56 mL of air at STP.The molecular mass of the compound is

6 g of a substance is dissolved in 100 g of water depresses the freezing point by 0.93^(@)C . The molecular mass of the substance will be: ( K_(f) for water = 1.86^(@)C /molal)