Answer : (a)

Molar mass of C2H6 = 24+6 = 30g 60g of C2H6= 2 moles Burning of C2H6 equation is 2C2H6 + 7O2 → 4CO2 + 6H2O Thus 2 moles of C2H6 required 7 moles of O2 Now volume of 1 mole of gas at STP = 22.4 L Volume of 7 moles of O2= 22.4 × 7 L Answer: (c)

Atomic mass unit is one twelfth of mass of C-12. A mass of one mole C = Avogadro number × mass of one atom of C Answer: (d)

Dibasic acid liberate two H+ ions which is required for calculation of Normality For given solute molecular wt = 200 ∴ Equivalent weight = 200/2= 100 Normality is given by formula ∴ w = 1 g Answer: (c)

Mass of one mole of water or molecular mass =18g As volume = mass/density ∴ Volume of 1 mole of water = 18 ml One mole contains 6.023× 10–23 number of molecules ∴ Volume of one molecule of water = 18/(6.023×1023 )=3×10-23 ml As 1ml = 1cm3 ∴ Volume occupy by one molecule of water = 3.0 × 10–23 cm3 Answer : (d)

Minimum one atom Of Sc must be present in given compound Now 0.5%= 78.4 ∴ 100% = 15680 = 1.568×104 Answer: (a)

% oxygen = 100 - 38.71 – 9.67 = 61.62% C = 38.71 ∴ Number of atoms 38.71/12 = 3.22 H = 9.67 ∴ Number of atoms 9.67/1 = 9.67 Thus Hydrogen atoms are three times the carbon Option b is correct Answer: (b)

Reaction N2 + 3H2 → 2NH3 Hydrogen is 30 L thus requires 10 Nitrogen Hydrogen is limiting reagent Now only 50% of Hydrogen is used = 15 and Balance hydrogen =15L Therefore 5 L of Nitrogen used and 30-5 = 25 Balanced NH3 Produced = 10 L Answer: (b)

Mixing solutions are same M1V1 + M2V2 = M(V1+V2) 1×2.5+0.5×3 = M(2.5+3) M= 0.7272 =0.73M Answer: (c)

Molecular weight of CH3OH = 12+4+16 = 32g M=(Wt of solvant×1000)/(Molar wt×volume of olution in mL) 2=(w×1000)/(32×150) W= 9.6 g Answer: (c)

Thermal decomposition of KClO3 2KClO3 → 2KCl +3O2 Thus Two moles of KClO3 produces Three moles of oxygen 1 mole of KClO3 will produce 1.5moleof Oxygen Now when Al burn 4Al +3O2 → 2Al2O3 Al is in excess thus oxygen is limiting reagent Now 3 moles of Oxygen produce 2 moles of Al2O3 × 1.5 moles of Oxygen produce 1 mole of Al2O3 Answer : (a)

If 5 g is mixture H2 = 1g = 0.5 mole( as molecular weight of H2= 2g) O2 = 4g = 0.125 moles( as molecular weight of O2 = 32g) Thus ration 0.5:0.125 =4:1 Answer: (c)

Answer : (d)

Reaction is 2Mg + O2 → 2MgO Two moles of Mg reacts with one mole of oxygen to produce 2moles of MgO 1.0g of Mg = 1/24 moles = 4/96 0.56 g of O2 = 0.56/32 moles = 1.68/96 For 1.68/96 moles of oxygen Mg requires = 2(1.68/96) Mg moles taken is 4/96 which is more than complete reaction of O2 Excess moles Mg=4/96 -(2×1.68)/96=0.64/96 Excess weight of Mg= Excess mole × molecular wt Excess weight =0.64/96×24=0.16g Answer : (a)

Molecular mass of KH=39+1=40g 4g of = 0.1 mole One molecule of KH have 19+1= 20 electrons ∴ 0.1 mole of KH have = 0.1×20 =2 mole of electrons Molecular mass of H2O= 2(1)+16 =18g 1.8g =0.1 mole One molecule of H2O have 2(1)+8 = 10 electron 0.1 mole of H2O have = 0.1×10 = 1 mole electrons The total number of electrons in 4 g of KH to that in 1.8 g of H2O is double Answer : (a)

MgCO3 → MgO + CO2 One mole of MgCO3 gives one mole each of MgO and CO2 Molecular weight of MgCO3 = 24+12+3(16) = 84g Molecular weight of MgO = 24+16 = 40g 40g of MgO = 84 MgCO3 ∴ 8 g of MgO = (84/5) g MgCO3 = 16.8 g But sample is 20g 20g sample = 16.8 MgCO3 100g same = 16.8 × 5 = 84 MgCO3 Answer : (b)

To make 0.1 M of H2SO4 Molecular weight of H2SO4 is 98 g ∴ 19.6 g of sulphuric acid require to make 1L of 0.2M solution As purity is 98% thus acid required will be =20 g Now density is 1.8gmL-1 therefore weight of 1000mL acid = 1800 g ∴ 20g = 11.11 mL Alternate method you can use formula for M if density and %w/w given M=(%w/w × density ×10)/(molecular mass) Then use M1V1 = M2V2 Answer (d)