Home
Class
CHEMISTRY
विलयन Solution | Chap 02 | L 02| 12th NE...

विलयन Solution | Chap 02 | L 02| 12th NEET 2021-22 | New Regular Course | Chemistry | Vikram Singh

Promotional Banner

Similar Questions

Explore conceptually related problems

A buffer solution 0.04 M in Na_(2)HPO_(4) and 0.02 in Na_(3)PO_(4) is prepared. The electrolytic oxidation of 1.0 milli-mole of the organic compound RNHOH is carried out in 100 mL of the buffer. The reaction is RNHOH+H_(2)OrarrRNO_(2)+4H^(+)+4e^(-) The approximate pH of solution after the oxidation is complete is : [Given : for H_(3)PO_(4),pK_(a1)=2.2,pK_(a2)=7.20,pK_(a3)=12]

A current of 965A is passed for 1 s through 1L solution of 0.02N NiSO_(4) using Ni electrodes. What is the new concentration of NiSO_(4) ?

In chemistry, 'mole' is an essential tool for the chemical calculations. It is a basic SI unit adopted by the 14^(th) general conference on weights and measurements in 1971. A mole contains as many elementary particles as the number of atoms present in 12 g of ""^(12)C . 1 mole of a gas at STP occupies 22.4 litre volume. Molar volume of solids and liquids is not definite. Molar mass of a substance is also called gram-atomic mass or gram molecular mass. The virtual meaning of mole is plenty, heap or the collection of large numbers. 1 mole of a substance contains 6.022 xx 10^(23) elementary particles like atom or molecule. Atomic mass unit (amu) is the unit of atomic mass, e.g., atomic mass of single carbon is 12 amu. XL N_(2) , gas at STP contains 3xx10^(22) molecules. The number of molecules in x L ozone at STP will be

According to the Avogadro's law , equal number of moles of gases occupy the same volume at identical conditions of temperature and pressure. Even if we have a mixture of non-reacting gases then Avogadro's law is still obeyed by assuming mixture as a new gas. Now let us assume air to consist of 80% by volume of nitrogen (N_(2)) and 20% by volume of oxygen (O_(2)) . If air is taken at STP , then its 1 mol would occupy 22.4 L . 1 mol of air would contain 0.8 mol of N_(2) and 0.2 mol of O_(2) hence the mole fraction of N_(2) and O_(2) are given by X_(n_(2))=0.8, X_(o_(2))=0.2 Density of air at STP is:

50 mL of an aqueous solution of glucose C_(6)H_(12)o_(6) (Molar mass : 180 g/mol) contains 6.02 xx10^(22) molecules. The concentration of the solution will be?