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HARDMCQ SINGLEJEE Advanced ChemistryPhysical ChemistryMole _ Equivalent Concept

5 mL of N-HCl, 20 mL of N/2 H₂SO₄ milli equivalent and 30 mL of N/3 HNO₃ are mixed together and the volume is made to 1L. The normality of the resulting solution is

HARDMCQ SINGLEJEE Advanced ChemistryPhysical ChemistryMole _ Equivalent Concept

Assertion :Pure water obtained from different states of India always contains hydrogen and oxygen in the ratio of 1 : 8 by mass. Reason: Total mass of reactants and products during chemical change is always the same.

HARDMCQ SINGLEJEE Advanced ChemistryPhysical ChemistryMole _ Equivalent Concept

Assertion :Both  12g  of  carbon  and  27g  of aluminium contain 6.023 × 1023 atoms. Reason :Molar  mass  of  an  element  contains Avogadro number of atoms.

HARDMCQ SINGLEJEE Advanced ChemistryPhysical ChemistryMole _ Equivalent Concept

Assertion: Equivalent weight of a base molecular weight = acidity Reason: Acidity is the number of replaceable hydrogen atom in one molecule of the base.

HARDMCQ SINGLEJEE Advanced ChemistryPhysical ChemistryMole _ Equivalent Concept

Assertion :  Both  138  g  of  K₂CO₃  and  12  g  of carbon have same number of carbon atoms. Reason :  Both  contains  1  g atom of  carbon  which contains 6.022 × 1023 carbon atoms.

HARDMCQ SINGLEJEE Advanced ChemistryPhysical ChemistryMole _ Equivalent Concept

Volume of acid that contains 63g pure acid is.

HARDMCQ SINGLEJEE Advanced ChemistryPhysical ChemistryMole _ Equivalent Concept

The total number of electrons present in 18 ml of water (density of water is 1 g ml–1) is

HARDMCQ SINGLEJEE Advanced ChemistryPhysical ChemistryMole _ Equivalent Concept

Assertion :Number of molecules present in SO₂ is twice the number of molecules present in O₂. Reason : Molecular mass of SO₂ is double to that of O₂.

HARDMCQ SINGLEJEE Advanced ChemistryPhysical ChemistryMole _ Equivalent Concept

A  solution  containing  12.0%  NaOH  by  mass  has  a density of 1.131 g/mL. What volume of this solution contains 5.00 mol of NaOH ?

HARDMCQ SINGLEJEE Advanced ChemistryPhysical ChemistryMole _ Equivalent Concept

A solution of NaOH is prepared by dissolving 4.0 g of  NaOH in 1  L of water. Calculate the volume of the HCl gas at STP that will neutralize 50 mL of this solution.

HARDMCQ SINGLEJEE Advanced ChemistryPhysical ChemistryMole _ Equivalent Concept

A  27.0  g  sample  of  an  unknown  hydrocarbon  was burned in excess O₂ to form 88g of CO₂ and 27g of H₂O.  What  is  possible  molecular  formula  of hydrocarbon ?

HARDMCQ SINGLEJEE Advanced ChemistryPhysical ChemistryMole _ Equivalent Concept

When a hydrate of Na₂CO₃ is heated until all the water is removed, it loses 54.3 per cent of its mass. The formula of the hydrate is

HARDMCQ SINGLEJEE Advanced ChemistryPhysical ChemistryMole _ Equivalent Concept

Volume of water required to make 1N solution from 2 mL conc. HNO₃.

HARDMCQ SINGLEJEE Advanced ChemistryPhysical ChemistryMole _ Equivalent Concept

If  we  consider  that  1/6,  in  place  of  1/12,  mass  of carbon  atom  is  taken  to  be  the  relative  atomic  mass unit, the mass of one mole of a substance will

HARDMCQ SINGLEJEE Advanced ChemistryPhysical ChemistryMole _ Equivalent Concept

In a compound C, H and N atoms are present as 9%, 1%, 3.5% respectively. Molecular weight of compound is 108. Molecular formula of compound is (a) C H N (b) C H N (c) C H N (d) C H N

HARDMCQ SINGLEJEE Advanced ChemistryPhysical ChemistryMole _ Equivalent Concept

Number  of  atoms  in  560g  of  Fe  (atomic  mass 56 g mol–1) is

HARDMCQ SINGLEJEE Advanced ChemistryPhysical ChemistryMole _ Equivalent Concept

Calculate  the  analytical  molarity  of  Cl–  ion  in solution which is prepared by mixing 100 ml of 0.1 M NaCl and 400 ml of 0.01 M BaCl₂.

HARDMCQ SINGLEJEE Advanced ChemistryPhysical ChemistryMole _ Equivalent Concept

The  molarity  of  68  %  of  H₂SO₄  whose  density  is 1.84 g/cc is

HARDMCQ SINGLEJEE Advanced ChemistryPhysical ChemistryMole _ Equivalent Concept

HCl is 80% ionised in 0.01 M aqueous solution. The equilibrium molarity of HCl in the solution is

HARDMCQ SINGLEJEE Advanced ChemistryPhysical ChemistryMole _ Equivalent Concept

The following process has been used to obtain iodine from oil-field brines in California. NaI + AgNO₃ → AgI + NaNO₃ AgI + Fe  → FeI₂ + Ag FeI₂ + Cl₂ → FeCl₃ + I₂ If 381 kg of iodine is produced per hour then mass of A…

HARDMCQ SINGLEJEE Advanced ChemistryPhysical ChemistryMole _ Equivalent Concept

NX is produced by the following step of reactions M + X2 → M X2 MX2 + X2 → M3X8 M3X8 + N₂CO₃ → N X + CO₂ + M3O₄ How  much M (metal) is consumed to  produce  206 gm of NX. (Take At. wt of M = 56, N=23, X = 80)

HARDMCQ SINGLEJEE Advanced ChemistryPhysical ChemistryMole _ Equivalent Concept

100  mL  of  mixture  of  NaOH  and  Na₂SO₄  is neutralised  by  10  mL  of  0.5  M  H₂SO₄.  Hence, NaOH in 100 mL solution is

HARDMCQ SINGLEJEE Advanced ChemistryPhysical ChemistryMole _ Equivalent Concept

Assertion : If 30 mL of H₂ and 20 mL of O₂ react to form  water,  5  mL  of  H₂  is  left  at  the  end  of  the reaction Reason :H₂ is the limiting reagent.

HARDMCQ SINGLEJEE Advanced ChemistryPhysical ChemistryMole _ Equivalent Concept

Assertion :For  a  binary  solution  of  two  liquids,  A and  B,  with  the  knowledge  of  density  of  solution, molarity can be converted into molality. Reason: Molarity is defined in terms of volume and molality …

HARDMCQ SINGLEJEE Advanced ChemistryPhysical ChemistryMole _ Equivalent Concept

Assertion :  1mole  of  H₂SO₄  is  neutralised  by  2 moles  of  NaOH  but  1  equivalent  of  H₂SO₄  is neutralised by 1 equivalent of NaOH. Reason : Equivalent weight of H₂SO₄ is half of its molecular weight while e…

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