# NEET Chemistry Questions Solved

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How many grams of concentrated nitric acid solution should be used to prepare 250 mL of 2.0 M HNO3? The concentrated acid is 70% HNO3.

(a) 45.0 g conc. HNO3

(b) 90.0 g conc. HNO3

(c) 70.0 g conc. HNO3

(d) 54.0 g conc. HNO3

Concept Videos :-

#8 | Conc. of Solutions: II
#11 | Questions on Equivalent Concept
#12 | Normality of Mixed Solution

Concept Questions :-

Concentration Based Problem

(a) Given, molarity of solution=2

Volume of solution = 250mL=250/1000=1/4L

Molar mass of HNO3 = 1+14+3x16=63g mol-1

Molarity =

Weight of  HNO3 = molarity x molecular massx volume(L) = 2x63x1/4g=31.5g

It is the weight of 100%  HNO3

But the given acid is 70% HNO3

Its weight = 31.5x100/70g=45g

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Mole fraction of the solute in a 1.00 molal aqueous solution is

(a) 0.0177

(b) 0.0344

(c) 1.7700

(d) 0.1770

Concept Videos :-

#1 | Mole Concept
#2 | Questions on Mole Concept

Concept Questions :-

Moles, Atoms & Electrons

(a) 1.00 molal aqueous solution = 1.0 mole in 1000g water

nsolute = 1; Wsolvent =1000g

nsolvent = 1000/18 = 55.56

Xsolute$\frac{{n}_{solute}}{{n}_{solute}+{n}_{solvent}}$

Xsolute = 1/1+55.56 = 0.0177

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6.02x1020 molecules of urea are present in 100mL of its solution. The concentration of solution is

(a) 0.02M

(b) 0.01M

(c) 0.001M

(d) 0.1M

Concept Videos :-

#1 | Mole Concept
#2 | Questions on Mole Concept

Concept Questions :-

Moles, Atoms & Electrons

(b) Given, number of molecules of urea= 6.02x1020

Number of moles = 6.02x1020 /NA

=6.02x1020 /6.02x1023 =1x10-3 mol

Volume of the solution = 100mL = 100/1000L =0.1L

Concentration of urea solution (in mol L-1)

=1x10-3 /0.1 mol L-1

=1x10-2 mol L-1

=0.01 mol L-1

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Which has maximum number of molecules?

(a) 7g N

(b) 2g H

(c) 16g NO

(d) 16g O

Concept Videos :-

#1 | Mole Concept
#2 | Questions on Mole Concept

Concept Questions :-

Moles, Atoms & Electrons

(b) In 7g nitrogen, number of  molecules = $\frac{7.0}{28}$mol = 0.25 x NA molecules

where, NA = Avogadro number = 6.023 x 1023

In 2g of H2= 2/2 mol =1x NA molecules

In 16 g of  NO2= 16/46 mol = 0.348 x NA molecules

In 16 g of O2 = 16/32 mol = 0.5 x NA molecules

Hence, maximum number of molecules are present in 2g of H2.

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The number of moles of KMnO4 reduced by one mole of KI in alkaline medium is

(a) one fifth

(b) five

(c) one

(d) two

Concept Videos :-

#5 | Stoichiometry
#6 | Questions on POAC
#13 | Calculation of % composition & % Purity

Concept Questions :-

Equation Based Problem

(d)  as per the balanced equation, one mole of KI in basic medium need 2 moles of KMnO4

## Video Solution:

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The number of moles of oxygen in 1L of air containing 21% oxygen by volume, under standard conditions, is

(a) 0.0093 mole

(b) 2.10 moles

(c) 0.186 mole

(d) 0.21 mole

Concept Videos :-

#1 | Mole Concept
#2 | Questions on Mole Concept

Concept Questions :-

Moles, Atoms & Electrons

(a) Volume oxygen in 1L of air = 21/100x1000 = 210mL

22400 mL volume at STP is occupied by oxygen = 1mole

Therefore, number of moles occupied by 210mL = 210/22400 = 0.0093mol

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What volume of  oxygen gas (O2) measured at $0°\mathrm{C}$ and 1 atm, is needed to burn completely 1L of  propane gas (C3H8) measured under the same conditions?

(a) 7L

(b) 6L

(c) 5L

(d) 10L

Concept Videos :-

#5 | Stoichiometry
#6 | Questions on POAC
#13 | Calculation of % composition & % Purity

Concept Questions :-

Equation Based Problem

(c) C3H8 + 5O2  $\to$3CO2+ 4H2O

22.4L   5x22.4L

For the combustion of 22.4L propane, oxygen required = 5 x 22.4L

For  the combustion of 1L of propane oxygen required = 5x22.4/22.4L

=5L

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1 g of magnesium is burnt with 0.56g of oxygen in a closed vessel. Which reactant is left in excess and how much?

(At. weight of Mg = 24, O=16)

(a) Mg, 0.16g

(b) O2, 0.16g

(c) Mg, 0.44g

(d) O2, 0.28g

Concept Videos :-

#5 | Stoichiometry
#6 | Questions on POAC
#13 | Calculation of % composition & % Purity

Concept Questions :-

Equation Based Problem

(a) The balanced chemical equation is

From the above equation, it is clear that, 24g of Mg reacts with 16g of O2 .

Thus,1.0 g of Mg reacts with 16/24g of O2  = 0.67g of O2

But only 0.56g of O2 is available which is less than 0.67g. Thus,O2 is the limiting reagent. Further, 16g of O2 reacts with 24g of Mg.

0.56g of O2 will react with Mg = 24/16x0.56 = 0.84g

Amount of Mg left unreacted = (1.0-0.84)g Mg = 0.16g of Mg

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Percentage of Se in peroxidase anhydrase enzyme is 0.5% by weight (at. weight = 78.4), then minimum molecular weight of peroxidase anhydrase enzyme is

(a) 1.568 x 10

(b) 15.68

(c) 2.168 x 10

(d) 1.568 x 10

Concept Videos :-

#1 | Mole Concept
#2 | Questions on Mole Concept

Concept Questions :-

Moles, Atoms & Electrons

(d) Suppose the molecular weight of enzyme =  x

0.5% by weight means in 100g of enzyme weight of Se=0.5g

In xg of enzyme weight of Se = $\frac{.5}{100}×x\phantom{\rule{0ex}{0ex}}$

Hence, 78.4 =

x = 15680 = 1.568x104

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Number of moles of Mn${\mathrm{O}}_{4}^{-}$ required to oxidise one mole of ferrous oxalate completely in acidic medium will be

(a) 0.6 mole

(b) 0.4 mole

(c) 7.5 moles

(d) 0.2 mole

Concept Videos :-

#5 | Stoichiometry
#6 | Questions on POAC
#13 | Calculation of % composition & % Purity

Concept Questions :-

Equation Based Problem

(1) 0.6

10FeC2O4 + 6KMnO4 + 24H2SO----> 3K2SO4+ 6MnSO+  5Fe2(SO4)3 + 24H2O + 20CO2

So we see that  6 moles of KMnO4 is required to oxidize 10 moles of  FeC2O4  Then, 1 mole of FeC2O4 would be oxidized by = ?    6/10 = 0.6

## Video Solution:

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