Questions: Some Basic Concepts of Chemistry

Q1 Define the mole. How many atoms are present in 0.5 mol of helium?

Answer (English): The mole is the SI unit of the amount of substance. One mole contains exactly 6.022 x 10^23 elementary entities, the number called the Avogadro constant. The entity must be stated: atoms, molecules, ions or formula units. For 0.5 mol of helium, each mole contains 6.022 x 10^23 atoms, so 0.5 mol contains 0.5 x 6.022 x 10^23 = 3.011 x 10^23 helium atoms. The answer carries the unit atoms, which matters because the same number of moles of a diatomic gas would give twice as many atoms.

Hinglish: Mole amount of substance ka SI unit hai aur ek mole mein 6.022 x 10^23 entities hoti hain, jise Avogadro constant kehte hain. Entity batana zaroori hai: atom, molecule ya ion. Helium ke 0.5 mol mein 0.5 x 6.022 x 10^23 = 3.011 x 10^23 atoms honge, kyunki helium monatomic hai.

Q2 What is the difference between molarity and molality? Why does molarity change with temperature?

Answer (English): Molarity is the number of moles of solute in one litre of solution, so it uses the total solution volume. Molality is the number of moles of solute in one kilogram of solvent, so it uses the solvent mass. Molarity changes with temperature because the volume of a liquid expands or contracts as temperature changes, while the mass of the solvent stays fixed. Molality therefore remains constant over a temperature range and is preferred when the temperature is not fixed.

Hinglish: Molarity = solute ke moles per litre of solution, jabki molality = solute ke moles per kg solvent. Temperature badhne par solution ka volume badalta hai, isliye molarity badalti hai, lekin solvent ka mass same rehta hai, isliye molality fixed rehti hai. Exam mein dono ke units aur denominators clear rakho.

Q3 State the law of multiple proportions and illustrate it with carbon monoxide and carbon dioxide.

Answer (English): If two elements can form more than one compound, fix the mass of the first element. The masses of the second element that pair with it then stand in a ratio of simple integers. Take 12 g of carbon as the fixed mass. In carbon monoxide that carbon combines with 16 g of oxygen; in carbon dioxide it combines with 32 g of oxygen. The two oxygen masses are in the ratio 1 to 2, so the law holds for these two compounds.

Hinglish: Law of multiple proportions kehta hai ki jab do elements ek se zyada compound banate hain, to fixed mass ke saath combine hone wala doosre element ka mass small whole number ratio mein hota hai. CO mein 12 g carbon 16 g oxygen se, aur CO2 mein 32 g oxygen se milta hai; 16:32 = 1:2, jo law ko satisfy karta hai.

Q4 Calculate the number of molecules in 11.2 g of carbon monoxide. What is the number of atoms in the same sample?

Answer (English): The molar mass of CO is 28 g mol^-1, so 11.2 g is 11.2 divided by 28, which is 0.40 mol. The number of molecules is 0.40 x 6.022 x 10^23 = 2.409 x 10^23. Each carbon monoxide molecule has two atoms, one of carbon and one of oxygen, so the number of atoms is twice that value, 4.818 x 10^23. Both products are obtained by converting mass to moles first and only then multiplying by the Avogadro constant.

Hinglish: CO ka molar mass 28 g/mol hai, isliye 11.2 g = 0.40 mol. Molecules = 0.40 x 6.022 x 10^23 = 2.409 x 10^23. Har CO molecule mein do atoms (C aur O) hote hain, isliye atoms ki sankhya double hogi, 4.818 x 10^23.

Q5 What are significant figures? Give two rules for counting them.

Answer (English): Significant figures are the digits in a measurement that are known with some confidence; they include the measured digits and one estimated digit. A rule for counting is that all non-zero digits are significant, so 234 has three significant figures. A second rule is that zeros between non-zero digits are significant, so 208 has three. Leading zeros are not significant, which is why 0.0025 has two significant figures, and trailing zeros count only when the number contains a decimal point or a marked decimal.

Hinglish: Significant figures measurement ke bharosemand digits hote hain, jisme last digit estimated hota hai. Non-zero digits hamesha significant hote hain; beech waale zeros bhi count karte hain (208 mein 3 sig figs). Leading zeros count nahi karte (0.0025 mein 2 sig figs), aur decimal ke baad waale trailing zeros count karte hain.

Q6 A compound contains 40.0 percent carbon, 6.7 percent hydrogen and 53.3 percent oxygen by mass. Find its empirical formula.

Answer (English): Take 100 g of the compound, so there are 40.0 g of carbon, 6.7 g of hydrogen and 53.3 g of oxygen. Divide each mass by the atomic mass: carbon gives 40.0 divided by 12.0, which is 3.33; hydrogen gives 6.7 divided by 1.0, which is 6.7; oxygen gives 53.3 divided by 16.0, which is 3.33. Divide all three by the smallest quotient 3.33: carbon 1, hydrogen 2, oxygen 1. The empirical formula is CH2O, which is the formula of formaldehyde.

Hinglish: 100 g sample lo: C = 40 g, H = 6.7 g, O = 53.3 g. Atomic mass se divide karo: C = 3.33, H = 6.7, O = 3.33. Sabko 3.33 se divide karo to C : H : O = 1 : 2 : 1, isliye empirical formula CH2O hai.

Q7 In the reaction 2Al + 3Cl2 giving 2AlCl3, 5.4 g of aluminium is treated with 7.1 g of chlorine. Find the limiting reagent and the mass of aluminium chloride formed.

Answer (English): The molar mass of aluminium is 27 g mol^-1, so 5.4 g is 0.20 mol. The molar mass of chlorine is 71 g mol^-1, so 7.1 g is 0.10 mol. Divide each by its coefficient: aluminium gives 0.20 divided by 2, which is 0.10; chlorine gives 0.10 divided by 3, which is 0.033. Chlorine has the smaller quotient, so chlorine is the limiting reagent. Chlorine gives two moles of aluminium chloride per three moles of chlorine, so the product is 0.10 x 2/3 = 0.067 mol. Its mass is 0.067 x 133.3 = 8.9 g.

Hinglish: Al ke 5.4 g = 0.20 mol, Cl2 ke 7.1 g = 0.10 mol. Coefficient se divide karo: Al = 0.20/2 = 0.10, Cl2 = 0.10/3 = 0.033. Chlorine limiting reagent hai. Product = 0.10 x 2/3 = 0.067 mol, aur mass = 0.067 x 133.3 = 8.9 g AlCl3.

Q8 What volume of oxygen at STP is needed to burn 4.0 g of methane completely?

Answer (English): The reaction is CH4 + 2O2 giving CO2 + 2H2O, so one mole of methane needs two moles of oxygen. Methane has a molar mass of 16 g mol^-1, so 4.0 g is 0.25 mol. The oxygen needed is 0.50 mol. At STP the molar volume is 22.7 L mol^-1 according to the rationalised NCERT text, so the volume is 0.50 x 22.7 = 11.35 L. JEE papers that use the older convention of 22.4 L mol^-1 would report about 11.2 L, and the working is otherwise the same.

Hinglish: CH4 + 2O2 → CO2 + 2H2O, isliye 1 mol methane ke liye 2 mol oxygen chahiye. 4.0 g methane = 0.25 mol, to oxygen 0.50 mol hoga. NCERT ke 22.7 L/mol se volume = 0.50 x 22.7 = 11.35 L; JEE mein 22.4 L/mol use hota hai to 11.2 L aata hai.

Q9 Explain the difference between empirical formula and molecular formula with an example.

Answer (English): The empirical formula gives the simplest whole number ratio of atoms in a compound, while the molecular formula gives the actual number of each kind of atom in one molecule. Benzene has the empirical formula CH and the molecular formula C6H6, so six empirical units make one molecule. The molecular formula is obtained by multiplying the empirical formula by a whole number n, where n is the molar mass divided by the empirical formula mass.

Hinglish: Empirical formula simplest whole number ratio deta hai, aur molecular formula ek molecule mein atoms ki actual sankhya deta hai. Benzene ka empirical formula CH aur molecular formula C6H6 hai. Molecular formula = empirical formula x n, jahan n = molar mass / empirical formula mass.

Q10 Why is the mass of a sample conserved during a chemical reaction, and how does this support the mole concept?

Answer (English): A chemical reaction only rearranges atoms into new combinations; no atom is created or destroyed. The mass of the products therefore equals the mass of the reactants. That is why a balanced equation can be read as a statement about moles: the coefficients give the ratio in which particles react, and multiplying by the molar mass gives the ratio in which masses react.

Hinglish: Chemical reaction mein atoms sirf naye combination banate hain, koi atom na banta hai na khatam hota hai, isliye products ka mass reactants ke mass ke barabar hota hai. Balanced equation ke coefficients particles ka mole ratio dete hain, aur molar mass se wahi ratio masses mein badal jaata hai. Isliye conservation of mass aur mole concept ek hi bookkeeping ke do roop hain.

That closes the chapter questions. Try the quick quiz below before moving on to the previous year questions.