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Q. What is the difference between molarity and normality?

Molarity based on vol. Of solvent in 100litre of solu. While normality is gram molecular mass

50 mL of 0.5 M oxalic acid is needed to neutralize 25 mL of sodium hydroxide solution. The amount of $NaOH$ in 50 mL of the given sodium hydroxide solution is :

• Option 1)

20 g

• Option 2)

4 g

• Option 3)

10 g

• Option 4)

40 g

Molar Mass - The mass of one mole of a substance in grams is called its molar mass. - wherein  Molar mass of water = 18 g mol-1       Molarity -  Molarity (M) = (Number of moles of solute)/(volume of solution in litres)   - wherein It is defined as the number of moles of the solute in 1 litre of the solution.       Stoichiometry - Stoichiometry deals with measurements of reactants and...

0.27 g of a long chain fatty acid was dissolved in 100 cm3 of hexane. 10 mL of this solution was added dropwise to the surface of water in a round watch glass. Hexane evaporates and a monolayer is formed. The distance from edge to centre of the watch glass is 10 cm. What is the height of the monolayer?

[Density of fatty acid=0.9 g cm-3 , $\pi=3$]

• Option 1)

10-6 m

• Option 2)

10-8 m

• Option 3)

10-2 m

• Option 4)

10-4 m

Density - Density is a measure of mass per unit  volume. The average density of an object is equals to its total mass divided by its total volume. An object made from a comparatively dense material (such as iron) will have less volume than an object of equal mass made from some less dense substance (such as water). To find the density of any object, we need to know the Mass (grams) of the...

The strength of 11.2 volume solution of $H_{2}O_{2}$ is [ Given that molar mass of H=1 g mol-1 and O=16 g mol-1]

• Option 1)

13.6%

• Option 2)

3.4%

• Option 3)

34%

• Option 4)

1.7%

volume strength of H2O2 - Volume or percentage strength of hydrogen peroxide is a term to express concentration of H2O2 in terms of volumes of oxygen gas ,based on its decomposition to form water and oxygen. 2H2O2 = 2 H2O + O2 Suppose we have a solution of M moles of H2O2 in 1L solution From the reaction stoichiometry, 2 moles of H2O2 gives 1mol of oxygen or 22.4L at S.T.P 1mole of H2O2 =...

For a reaction,

$N_{2}(g)+3H_{2}(g)\rightarrow 2NH_{3}(g);\, \, \, \, \, \: \: \; \: \:$

identify dihydrogen $\left ( H_{2} \right )$ as a limiting reagent in the following reaction mixtures.

• Option 1)

$56\, g\: of\: N_{2}\: +\: 10\, g\: of\: H_{2}$

• Option 2)

$35\: g\: of\: N_{2}\: +\: 8\: g\: of\: H_{2}$

• Option 3)

$28\: g\: of\: N_{2}\: +\: 6\: g\: of\: H_{2}$

• Option 4)

$14\: g\: of\: N_{2}\: +\: 4\: g\: of\: H_{2}$

Concept of limiting reagent and excess reagent - The reactant which gets consumed and thus limits the amount of product formed is called the limiting reagent. - wherein  CH4 (g) + 2O2 (g) → CO2 (g) + 2 H2O (g) ; In this reaction, if 1 mole of methane and 1 mole of oxygen are taken then oxygen would be limiting reagent.       For :- Identifying  as a limiting reagent                     ...

At 300 K and 1 atmsopheric pressure, 10 mL of a hydrocarbon required 55 mL of $O_{2}$

for complete combustion, and 40 mL of $CO_{2}$  is formed. The formula of the hydrocarbon is:

• Option 1)

$C_{4}H_{10}$

• Option 2)

$C_{4}H_{6}$

• Option 3)

$C_{4}H_{7}Cl$

• Option 4)

$C_{4}H_{8}$

The eqn of hydrocarbon can be written as :  Since , 10mL of  produces 40mL of  &               1mL of  produces  mL of                         Now,          10mL of   requires 55mL of          1mL of   requires mL of  =>  =>  => => So, option (2) is correct.Option 1)Option 2)Option 3)Option 4)

$5\: moles$ of $AB_2$  weigh $125\times 10^{-3}kg$ and $10\: moles$ of $A_2B_2$ weigh $300\times 10^{-3}kg.$ The molar mass of $A(M_A)$ and molar mass of $B(M_B)$ in $kg\: mol^{-1}$ are :

• Option 1)

$M_A=50\times 10^{-3}\: \: and\: \: M_B=25\times 10^{-3}$

• Option 2)

$M_A=25\times 10^{-3}\: \: and\: \: M_B=50\times 10^{-3}$

• Option 3)

$M_A=5\times 10^{-3}\: \: and\: \: M_B=10\times 10^{-3}$

• Option 4)

$M_A=10\times 10^{-3}\: \: and\: \: M_B=5\times 10^{-3}$

Molar Mass - The mass of one mole of a substance in grams is called its molar mass. - wherein  Molar mass of water = 18 g mol-1       Mole Concept - One mole is the amount of a substance that contains as many particles or entities as there are atoms in exactly 12 g (or 0.012 kg) of the 12C isotope. - wherein one mole = 6.0221367 X 1023          Let atomic mass of   & atomic mass of...

The mole fraction of a solvent in aqueous solution of a solute is $0.8.$ The molality $\left ( in\: \: \: mol\: Kg^{-1} \right )$  of the aqueous solution is :

• Option 1)

$13.88\times 10^{-2}$

• Option 2)

$13.88\times 10^{-3}$

• Option 3)

$13.88\times 10^{-1}$

• Option 4)

$13.88$

Mole Fraction - It is ratio of moles of solute or moles of solvent to moles of solution. - wherein If a substance ‘A’ dissolves in substance ‘B’ and their number of moles are  and  respectively; then the mole fractions of A and B are given as Mole fraction of A = (number of moles of A)/(number of moles of solution ) = /(+ )         We have,                        Option 1)Option...

The minimum amount of $O_{2}\left ( g \right )$ consumed per gram of reactant is for the reaction :

(Given atomic mass :$Fe=56,$$O=16$$Mg=24,P=31,C=12,H=1$)

• Option 1)

$4Fe\left ( s \right )+3O_{2}\left ( g \right )\rightarrow 2Fe_{2}O_{3}\left ( s \right )$



• Option 2)

$P_{4}\left ( s \right )+5O_{2}\left ( g \right )\rightarrow P_{4}O_{10}\left ( s \right )$

• Option 3)

$C_{3}H_{8}\left ( g \right )+5O_{2}\left ( g \right )\rightarrow 3CO_{2}\left ( g \right )+4H_{2}O\left ( I \right )$

• Option 4)

$2Mg\left ( s \right )+O_{2}\left ( g \right )\rightarrow 2MgO\left ( s \right )$

mole  required for  mole         per gram  ,  required (minimum) per gram  required=  per gram  required=  per gram  required =     Option 1)         Option 2)Option 3) Option 4)

25g of an unknown hydrocarbon upon burning produces 88 g of $CO_{2}$ and 9 g of $H_{2}O$. This unknown hydrocarbon contains:

• Option 1)

20 g of carbon and 5 g of hydrogen

• Option 2)

22 g of carbon and 3 g of hydrogen

• Option 3)

24 g of carbon and 1 g of hydrogen

• Option 4)

18 g of carbon and 7 g of hydrogen

weight of carbon  Mole of water  weight of water     Option 1)20 g of carbon and 5 g of hydrogen Option 2)22 g of carbon and 3 g of hydrogen Option 3)24 g of carbon and 1 g of hydrogen Option 4)18 g of carbon and 7 g of hydrogen

What would be the molality of $20\:^{o}/_{o}$  (mass/mass) aqueous solution of KI?

( molar mass of KI =$166\:g\:\:mol^{-1}$  )

• Option 1)

$1.08$

• Option 2)

$1.35$

• Option 3)

$1.48$

• Option 4)

$1.51$

Molality - Molality (m) = (number of moles of solute)/(mass of solvent in kg)   - wherein It is defined as the number of moles of solute present in 1 kg of solvent. It is denoted by m.       Mass percent of solution - Mass percent = ((mass of solute)/(mass of solution) ) X 100   - wherein  it is mass of solute present in 100 gram solution.            (mass/mass) = 20 gm is present in 100 gm...

$10 \:\:mL$ of  $1\:\: mM$ surfactant solution forms a monolayer covering $0.24\:\:cm^{2}$ on a polar substrate . If the polar head is approximated as a cube , what is its edge length ?

• Option 1)

$1.0\:pm$

• Option 2)

$2.0\:pm$

• Option 3)

$0.1\:nm$

• Option 4)

$2.0\:nm$

Mole Concept - One mole is the amount of a substance that contains as many particles or entities as there are atoms in exactly 12 g (or 0.012 kg) of the 12C isotope. - wherein one mole = 6.0221367 X 1023          Number of Moles - No of moles = given mass of substance/ molar mass of substance -             given  so and     given     Option 1) Option 2) Option 3) Option 4)

The percentage composition of carbon by mole in methane is:

• Option 1)

75%

• Option 2)

80%

• Option 3)

25%

• Option 4)

20%

mole % of 'C' in CH4 =                                      =  For 1 mole of C, there are 4 moles of hydrogen  and  %Option 1)75%Option 2)80%Option 3)25%Option 4)20%

In order to oxidise a mixture of one mole of each of $FeC_{2}O_{4},Fe_{2}(C_{2}O_{4})_{3},FeSO_{4}$ and $Fe_{2}(SO_{4})_{3}$ in acidic medium, the number of moles of $KMnO_{4}$ required is :

• Option 1)

$1.5$

• Option 2)

$3$

• Option 3)

$1$

• Option 4)

$2$

is the n=factor of and not In this compound, carbon is in oxidation state and is in oxidation state. In acidic medium , is converted to and is  converted to . So oxidation number of carbon changes from As, there are two carbon atoms, the net change in Number is and adding to this, the increase in oxidation state of iron, the final increase in Number is . Hence n-factor of is . ...
As we learnt in Mole Fraction - It is the ratio of moles of solute or moles of solvent to moles of the solution. - wherein If a substance ‘A’ dissolves in substance ‘B’ and their number of moles are  and  respectively; then the mole fractions of A and B are given as Mole fraction of A = (number of moles of A)/(number of moles of solution ) = /(+ )    At   ,                                      ...
Moles of urea = Number of molecules/Avogadro number => 6.02x10^20/6.02x10^23 => 10^-3 moles Now, concentration or molarity = Number of moles/ volume in litre => M = (10^-3/100)*1000 => M = 10^-2 M

Calculate the percentage composition in terms of mass of solution abtained by mixing 400g of 25% soln and 300g of 30% of soln?

Mass of component in first solution = 25% of 400 gram  = 25*400/100 = 100 gram Mass of component in second solution   = 30 % of 300 gram = 30*300/100 = 90 gram   Total mass of solution = 300+400 = 700 gram Total mass of component  = 100+ 90  = 190 grams   Mass % of final solution     = mass of component * 100 / total mass                                          =190*100/700  = 27.14% Total of...
As we have learnt   Molecular Formula - The molecular formula shows the exact number of different types of atoms present in a molecule of a compound. - wherein For glucose, empirical formula is CH2O .its molar mass is 180 gram. n = molar mass/empirical formula mass = 180/30=6 Hence molecular formula= C6H12O6    Molecular mass    Molecular Formula = Molecular mass
As we learnt Dalton's Atomic Theory - Chemical reactions involve reorganization of atoms. Atoms are neither created nor destroyed in a chemical reaction. - wherein The number of atoms on both sides of the equation is the same By Dalton's Atomic Theory, the atoms are neither created nor destroyed  moles of      molar mass of   = 513g % purity =   Solution is B Option 1) 25 This is...
As we Know Dalton's Atomic Theory: Chemical Reactions involve reorganization of atoms. Atoms are neither created nor destroyed in a chemical reaction. - wherein The number of atoms on both sides of the equation is the same By Dalton's Atomic Theory, the atoms are neither created nor destroyed  moles of      molar mass of   = 513g % purity =   Hence Correct answer is 50 %.
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