Given 98%H2SO4(w/v) Thus 100 ml ofH2SO4 have 98 g of H2SO4 or 1 mole of H2SO4 Now 100 mL have 1 mole 1L = 10 moles Therefore answer is 10M Answer : (c)

Mass of One molecule of Oxygen = 32/NA Mass of One atom of Nitrogen = 14/NA Mass of 10-10 g molecular weight of oxygen = 32× 10-10 g Mass of 10-10 g atomic weight of copper = 64× 10-10 g Answer : (a)

Molecular mass of KH=29+1=30g 3g 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 3 g of KH to that in 1.8 g of H2O is double Answer : (a)

4PH3 → P4 + 6H2( as Phosphorus is Tetra atomic molecules) 4 mL of PH3 gives 6mL of H2 ∴ 100 mL of PH2 gives 150mL of H2 Increase in volume = 150-100 =50 mL Answer : (a)

As density =1g/cc 55.5 moles of water occupy 1000 mL volume 55.5×6.023×1023 molecules = 1000 mL Volume 1 molecule = 0.00299×10-20 L Volume ∴ Volume of 1 molecule = 3×10-23 cm3 Answer : (d)

Using formula for average atomic mass Average mass=(200×90+199×8+202×2)/100 On calculating we get Average atomic mass = 199.08 ≈ 199amu Answer : (c)

Reaction is H2+ Cl2 → 2HCl One of each H2 and C2 produce 2 moles of HCl 22.4 L of H2= 1 mole 11.2 Lof Cl2 = 0.5 mole As Cl2 is only 0.5 mole ∴HCl produced will be 1 mole and Hydrogen of 0.5 mole will remain unreacted Answer : (a)

Answer : (a)

Answer : (a)

Reaction is H2SO4 + 2NaOH → Na2SO4+ 2H2O One mole of H2SO4 reacts with 2 moles of NaOH Since both the solution have same concentration 100 ml of H2SO4 will neutralize 200 mLof NaOH Thus 300 mL of NaOH of 0.1M will remain reacted =0.03mole of NaOH But after adding both the reactant volume of solution= 100+ 500 = 600 mL And mixture have 0.03 mole of NaOH ∴molarity of NaOH = 600mL =0.03 mole 1000 mL = N/20 ∴ mixture will be basic, N/20 Answer: (b)

1 Å = 10-10 m Volume of virus = πr2h = π ×(7×10-10)2× 10×-10 cc = 1.54×10-27 m3 6.02× 10-2 cc = 1 g Or 6.02 × 10-8 m3 = 10-3 kg ∴ weight of 1.54 × 10-27 m- = w W is the weight of one virus ∴molecular weight M of virus = w ×NA Answer : (a)

Number of atoms in one mole of triatomic gas= 3 . Number of atoms = 3 × 0.1 × 6.022 × 1023 = 1.806 × 1023 Answer (b)

Density is0.93g/cm3 or 0.93g/mL Thus weight of 1mL NH3 solution is 0.93g Since NH3solution if of 18.6% by weight ∴ 18.6%of 0.93 g = 0.1728 g ≈ 0.17g/cm3 Answer(a)

Given 0.33% Fe of 100gHaemoglobin = 0.33g Molecular mass of Fe = 56 i.e. 56g = 1 mole ∴ 0.33 g = 0.33/56 moles = 5.9 ×10-3 ∴ number molecules of Fe = 5.9 ×10-3× 6.023×1023 = 0.03553 ×1023 Atoms Answer: (a)

One mole of gasses occupy equal volume irrespective of there atomic masses CH4 + 2O2 → CO2 + 2H2O Or one mole of CH4 requires two moles of oxygen Thus 1L of CH4 requires 2 L of oxygen Answer : (b)

As one CO3-2 ion have two unit charge ∴ 1 mole of CO3-2 ion havee charge = 2 × 96500 C = 1.93 × 105 C Charge on one mole electron = 1.602 × 10-19× 6.022 × 1023 =96478 = 96500 C Answer : (c)

Molality = moles per Kg of solvent Given 3 moles of NaOH and solvent is water (aqueous)= 1Kg= 55.5 Thus mole faction x x=55.5/(3+55.5)=0.9487 Answer: (d)

Reaction 2H2 (g) + O2 (g) → 2H2 O(l) Two moles of H2 and one mole of O2 produce two moles of water 10g of hydrogen=5 moles 64g of oxygen = 2 moles ∴oxygen is limiting reagent , Now 2 moles will produce 4 moles of water Answer (b)

Reaction C3H8 + 5O2 → 3CO2 + 4H2O Thus 1 L requires 5Lof O2 Answer: (d)

Both atoms D2O and H2O have same number of electron Molecular weight of D2O = 20g Molecular weight of H2O =18 g Thus total number of electrons in 2.0 g of D2O to that in 1.8 g of H2O are same as number of moles are same = 0.1 mole Each H2O have 10 electrons Thus number of electron =1mole Answer: (b)