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At 300 K and 1 atm, 15 mL of a gaseous hydrocarbon requires 375 mL air containing 20% O2 by volume for complete combustion.  After combustion, the gases occupy 330 mL.  Assuming that the water formed is in liquid form and the volumes were measured at the same temperature and pressure, the formula of the hydrocarbon is :
Option: 1  C4H8  
Option: 2  C4H10
Option: 3  C3H6
Option: 4  C3H8
 

Volume of N2  in air = 375 × 0.8 = 300 ml

Volume of O2 in air = 375 × 0.2 = 75 ml
 

C_{x}H_{y} +\left ( x +\frac{y}{4} \right )O_{2} \; \rightarrow \; xCO_{2}(g) + \frac{y}{2} H_{2}O(l)

15ml                15\left ( x +\frac{y}{4} \right )

  0                         0                            15x                 -

 

After combustion total volume

330 =V_{N_{2}} + V_{CO_{2}}

330 = 300 + 15x 

x = 2 

Volume of O2 used

15\left ( x +\frac{y}{4} \right ) = 75

\left ( x +\frac{y}{4} \right ) = 5

y = 12 

So hydrocarbon is = C2H12

None of the options matches it therefore it is a BONUS.

----------------------------------------------------------------------

Alternatively  Solution


 C_{x}H_{y} +\left ( x +\frac{y}{4} \right )O_{2} \; \rightarrow \; xCO_{2}(g) + \frac{y}{2} H_{2}O(l)

15ml              15\left ( x +\frac{y}{4} \right )

  0                         0                            15x                 -

Volume of O2 used

15\left ( x +\frac{y}{4} \right ) = 75

\left ( x +\frac{y}{4} \right ) = 5

If further information (i.e., 330 ml) is neglected, option (C3H8 ) only satisfy the above equation.

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Posted by

Ritika Jonwal

A solution is prepared by mixing 8.5 g of CH2Cl2 and 11.95 g of CHCl3.  If vapour pressure of CH2Cl2 and CHCl3 at 298 K are 415 and 200 mmHg respectively, the mole fraction of CHCl3 in vapour form is : (Molar mass of Cl=35.5 g mol−1)  
Option: 1  0.162
Option: 2 0.675
Option: 3 0.325
Option: 4  0.486  
 

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Posted by

vishal kumar

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At 300 K, the density of a certain gaseous molecule at 2 bar is double to that of dinitrogen (N2) at 4 bar. The molar mass of gaseous molecule is :
Option: 1 28 g mol−1
Option: 2 56 g mol−1
Option: 3  112 g mol−1
Option: 4  224 g mol−1  
 

 

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Posted by

vishal kumar

Among the following, the incorrect statement is :
Option: 1  At low pressure, real gases show ideal behaviour.
Option: 2  At very low temperature, real gases show ideal behaviour.
Option: 3  At very large volume, real gases show ideal behaviour.
Option: 4  At Boyle’s temperature, real gases show ideal behaviour.  
 

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Posted by

vishal kumar

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The most abundant elements by mass in the body of a healthy human adult are : Oxygen (61.4%); Carbon (22.9%), Hydrogen (10.0%); and Nitrogen (2.6%). The weight (in kg) which a 75 kg person would gain if all ^{1}H atoms are replaced by ^{2}H atoms is :  
Option: 1 7.5
Option: 2 10
Option: 3 15
Option: 4 37.5
 

Given that

Mass of the person = 75 kg

Mass of 1H1 present in person = 10% of 75 kg = 7.5 kg

Since Mass of 1H2 is double the Mass of 1H1

So, Mass of 1H2 will be in person = 2 X 7.5 kg =15 kg

Thus, increase in weight = 15 - 7.5 = 7.5 kg

Therefore, Option (1) is correct

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Posted by

vishal kumar

The correct order of the atomic radii of C, Cs, Al and S is :
Option: 1 S<C<Cs<Al
Option: 2 C<S<Al<Cs
Option: 3 C<S<Cs<Al
Option: 4 S<C<Al<Cs
 

 

Periodicity of atomic radius and ionic radius in period -

In a period from left to right the effective nuclear charge increases because the next electron fills in the same shell. So the atomic size decrease.

- wherein

Li>Be>B>C>N>O>F

 

 

 

Electronegativity and atomic radius -

The attraction between the outer electrons and the nucleus increases as the atomic radius decreases in a period.

- wherein

Electronegativity\propto\frac{1}{atomic\:radius}

 

 

 

Size of atom and ion in a group -

In a group moving from top to the bottom the number of shell increases.So the atomic size increases.

- wherein

Li<Na<K<Rb<Cs

As we know that

From Left to right in a period size decreases and when going down the group size increases

C< S< Al< Cs

Therefore, Option(2) is correct

  

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Posted by

Ritika Jonwal

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The IUPAC symbol for the element with atomic number 119 would be:
Option: 1 uue
Option: 2une
Option: 3 unh
Option: 4 uun

 

Nomenclature of elements with atomic number >100 -

The name is derived directly from the atomic number of the element using the following numerical roots:

0 = nil

1 = un

2 = bi

3 = tri

4 = quad

5 = pent

6 = hex

7 = sept

8 = oct

9 = enn

Eg:

 

Atomic number

Name

Symbol

101

Mendelevium (Unnilunium)

Md (Unu)

102

Nobelium (Unnilbium)

No (Unb)



 

-

uue

1  1  9

Un Un ennium

Therefore, Option(1) is correct.

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Posted by

Ritika Jonwal

When photon of energy 4.0eV strikes the surface of a metal A, the ejected photoelectrons have maximum kinetic energy T_A\: eV and de-Broglie wavelength \lambda _A. The maximum kinetic energy of photoelectrons liberated from another metal B by photon of energy 4.50eV is T_B = (T_A-1.5)eV. If the de-Broglie wavelength of these photoelectrons \lambda _B=2\lambda _A, then the work function(in eV) of metal B is: 
Option: 1 4
Option: 2 1.5
Option: 3 3
Option: 4 2

As we know \lambda =\frac{h}{\sqrt{2mT}}

where T=kinetic energy

So \frac{\lambda_A}{\lambda_B} = \sqrt{\frac{T_A}{T_B}}=\frac{1}{2}...(1)

Also Given \ \ T_B=T_A-1.5... (2)

Using equation (1) and (2)

T_A=2 \ eV , \ and \ T_B=1.5 \ eV

And using the Einstein equation

E_B=T_B+W_B\\ \Rightarrow W_B=E_B-T_B=(4.5-0.5)eV=4 \ eV

So the correct option is 1.

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Posted by

vishal kumar

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The relative strength of interionic/intermolecular forces in decreasing order is:
Option: 1 ion - dipole > dipole - dipole > ion -ion
Option: 2 ion - ion > ion - dipole > dipole-dipole
Option: 3ion - dipole > ion -ion > dipole-dipole
Option: 4dipole-dipole > ion - dipole >  ion -ion
 

The correct order of intermolecular forces is:

ion-ion > ion-dipole > dipole-dipole

Therefore, Option(2) is correct.

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Posted by

Kuldeep Maurya

NaClO_{3} is used, even in spacecrafts, to produce O_{2} . The daily consumption of pure O_{2} by a person is 492L at 1\; atm,300\; K. How much amount of NaClO_{3} , in grams, is required to produce O_{2} for the daily consumption of a person at 1\; atm,300\; K ? _______. NaClO_{3}(s)+Fe(s)\rightarrow O_{2}(g)+NaCl(s)+FeO(s) R=0.082\; L\; atm\; mol^{-1}K^{-1}
 

moles of NaClO_{3} = moles of O_{2}

\mathrm{moles\: of\: O_{2}\: =\: \frac{PV}{RT}\: =\: \frac{1\, x\, 492}{0.082\, x\, 300}\: =\: 20ml}

Mass of NaClO_{3}=20\times 106.5=2130g

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Posted by

Kuldeep Maurya

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