Consider the reaction

$$4 \mathrm{HNO}_{3}(1)+3 \mathrm{KCl}(\mathrm{s}) \rightarrow \mathrm{Cl}_{2}(\mathrm{~g})+\mathrm{NOCl}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})+3 \mathrm{KNO}_{3}(\mathrm{~s})$$

The amount of $$\mathrm{HNO}_{3}$$ required to produce $$110.0 \mathrm{~g}$$ of $$\mathrm{KNO}_{3}$$ is

(Given: Atomic masses of $$\mathrm{H}, \mathrm{O}, \mathrm{N}$$ and $$\mathrm{K}$$ are $$1,16,14$$ and 39, respectively.)

$$ \mathrm{N}_{2(\mathrm{~g})}+3 \mathrm{H}_{2(\mathrm{~g})} \rightleftharpoons 2 \mathrm{NH}_{3(\mathrm{~g})} $$

$$20 \mathrm{~g} \quad ~~~5 \mathrm{~g}$$

Consider the above reaction, the limiting reagent of the reaction and number of moles of $$\mathrm{NH}_{3}$$ formed respectively are :

$$250 \mathrm{~g}$$ solution of $$\mathrm{D}$$-glucose in water contains $$10.8 \%$$ of carbon by weight. The molality of the solution is nearest to

(Given: Atomic Weights are, $$\mathrm{H}, 1 \,\mathrm{u} ; \mathrm{C}, 12 \,\mathrm{u} ; \mathrm{O}, 16 \,\mathrm{u}$$)

In Carius method of estimation of halogen, $$0.45 \mathrm{~g}$$ of an organic compound gave $$0.36 \mathrm{~g}$$ of $$\mathrm{AgBr}$$. Find out the percentage of bromine in the compound.

(Molar masses : $$\mathrm{AgBr}=188 \mathrm{~g} \mathrm{~mol}^{-1} ; \mathrm{Br}=80 \mathrm{~g} \mathrm{~mol}^{-1}$$)