A compound X contains $$32 \%$$ of A, $$20 \%$$ of B and remaining percentage of C. Then, the empirical formula of $$\mathrm{X}$$ is :

(Given atomic masses of A=64 ; B=40 ; C=32 u)

The density of 1 M solution of a compound 'X' is 1.25 g mL$$^{-1}$$. The correct option for the molality of solution is (Molar mass of compound X = 85 g):

The right option for the mass of $$\mathrm{CO}_{2}$$ produced by heating $$20 \mathrm{~g}$$ of $$20 \%$$ pure limestone is

(Atomic mass of $$\mathrm{Ca}=40$$ )

$$\left[\mathrm{CaCO}_{3} \stackrel{1200 \mathrm{~K}}{\longrightarrow} \mathrm{CaO}+\mathrm{CO}_{2}\right]$$

The density of the solution is 2.15 g mL^$$-$$1, then mass of 2.5 mL solution in correct significant figures is :