Some Basic Conc...

One mole of any substance contains $$6.022\times 10^{23}$$ atoms/molecules. Number of molecules of $$H_2SO_4$$ present in $$100$$mL of $$0.02$$M $$H_2SO_4$$ solution is __________.

The weight of mixture containing HCl and $$H_2SO_4$$ is $$0.1$$g on treatment with an excess of an $$AgNO_3$$ solution, reacted with this acid mixture gives $$0.1435$$g of AgCl, weight percentage of $$H_2SO_4$$ in mixture is:

The percentage composition of $$Cr$$ in $${ K }_{ 2 }{ Cr }_{ 2 }{ O }_{ 7 }$$ is (Given molar mass of $${ K }_{ 2 }{ Cr }_{ 2 }{ O }_{ 7 }=294.0g$$ $$mol$$)

Calculate the silver ion concentration in $$0.2\ M$$ solution as $$Ag(NH_{3})_{2}NO_{3} \cdot K_{silver} = 32\times 10^{-9}$$.

In carbon disulphide $$(CS_{4})$$, the mass of sulphur in combination with $$3.0\ g$$ of carbon is

A solution containing 4.2 g of KOH and $$Ca(OH)_2$$ is neutralized by an acid. If it consumes 0.1 equivalent of acid, the % of KOH in the original sample is: 

Sodium bicarbonate, $$NaHCO_{3}$$, can be purified by dissolving it in hot water $$(60^{\circ}C)$$, filtering to remove insoluble impurities, cooling to $$0^{\circ}C$$ to precipitate solid $$NaHCO_{3}$$, and then filtering to remove the solid, leaving soluble impurities in solution. Any $$NaHCO_{3}$$ that remains in the solution is not recovered. The solubility of $$NaHCO_{3}$$ in hot water at $$60^{\circ}C$$ is $$164\ g/litre$$ and is $$69\ g/litre$$ in cold water at $$0^{\circ}C$$. What is the percentage yield of $$NaHCO_{3}$$, when it is purified by this method?

$$H_3AO_4$$ has 31.6% of "A" by mass. The atomic mass of the atom "A" is:

$$\text{0.80 g}$$ of impure $$(NH_4)_2SO_4$$ was boiled with $$100 ml$$. of a $$\text{0.2 N NaOH}$$ solution till all the $$NH_3(g)$$ evolved.The remaining solution was diluted to $$\text{250 mL.25 mL}$$ of this solution was neutralized using $$5 mL$$ of a $$\text{0.2 n H_2SO_4}$$ solution.The percentage purity of the $$(NH_4)_2SO_4$$ sample is 

Two organic compounds $$\underline {R}$$ and $$\underline {S}$$, both containing only $$C$$ and $$H$$ yield on analysis, the same percentage composition by weight.
$$C = (12/ 13) \times 100$$% and $$H = (1/13)\times 100$$%.
$$\underline {R}$$ decolourises $$Br_{2}$$ - water but $$\underline {S}$$ does not.