40 miligram diatomic volatile substance `(X_2)` is converted to vapour that displaced `4.92mL` of air at 1atm and 300k. Atomic mass of element X is nearly :

40 mg of gaseous substance ( X_(2) ) occupies 4.8 mL of volume at 1 atm and 27^(@) C. Atomic mass of element X is : (R:0.08 atm L/mole- k )

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

In Victor Meyer's experiment, 0.6 g of a volatile substance displaced 112 mL of air at S.T.P. Find the molecular mass of the substance.

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

In a Victor Meyer's determination of molecular mass, 0.15 g of a volatile substance displaced 31.64 mL of air at 25°C and 755 mm pressure. Calculate the molecular mass of the substance.

In determination by Victor Meyer method, 0.60 g of a volatile substance expelled 123 mL of air measured over water at 20^(@)C and 757.4 mm pressure. Find the molecular mass of the substance. Aqueous tension at 20^(@)C is 17.4 mm.

The vapour pressure of acetone at 20^(@)C is 185 torr. When 1.2 g of non-volatile substance was dissolved in 100g of acetone at 20^(@)C its vapour pressure was 183 torr. The molar mass (g mol^(-1)) of the substance is:

Calculate the compressibility factor for CO_(2) if one mole of it occupies 0.4 litre at 300K and 40 atm . Comment on the result:

In a Victor Meyer's determination, 0.23 g of a volatile substance displaced air which measured 112 mL at S.T.P. Calculate the vapour density and molecular weight of the substance (1 litre of H_2 at S.T.P. weighs 0.09 g).

The ratio of the mass of C-12 atom to that of an atom of element X (whose atomicity is four) is 1:9 . The molecular mass of element x is :-