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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 O₂used

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

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

y = 12

So hydrocarbon is = C₂H₁₂

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

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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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vishal kumar

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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

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

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$NaClO_{3}$ 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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Kuldeep Maurya

Which of the following compounds is likely to show both Frenkel abnd Schottky defects in its crystalline form ?
Option: 1 AgBr

Option: 2 CsCl

Option: 3 KBr

Option: 4 ZnS

2

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Posted by

Raj Singh

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A beam of electromagnetic radiation of intensity $6.4\times10^{-5} W/cm^2$ is comprised of wavelength $\lambda=310 nm$ . It falls normally on a metal (work function $\phi$ =2 eV) of surface area 1 cm². If one in 10³ photons ejects an electron, total number of electrons ejected in 1s is 10^x (hc=1240 eVnm, 1 eV=1.6 x 10^-19J), then x is :-
Option: 1 11
Option: 2 10
Option: 3 12
Option: 4 13

Energy in 1 second = $6.4 \times 10 ^{-5} \times \1 \ \times 1 = 6.4 \times 10 ^{-5} Joule$

$Energy (eV) = \frac{6.4 \times 10 ^{-5}}{1.6 \times 10^{-19}} = 4 \times 10^{14} eV$

$Energy \ of\ one\ photon = \frac{12400}{3100} = 4 \ eV$

$Number \ of \ photon \ = \frac{4 \times 10^{14}}{4} = 10^{14}$

$Number \ of \ electrons \ = \frac{10^{14}}{10^3} = 10^{11}$

$So, \ x = 11$

So the correct answer is option 1.

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Ritika Jonwal

An electron of mass m and magnitude of charge $\left | e \right |$ initially at rest gets accelerated by a constant electric field E.The rate of change of de-broglie wavelength of this electron at time t is ? ( ignoring relativistic effects is) :
Option: 1 $\frac{-h}{\left | e \right |Et^{2}}$
Option: 2   $\frac{\left | e \right |Et}{h}$
Option: 3   $-\frac{h}{\left | e \right |Et}$
Option: 4   $-\frac{h}{\left | e \right |E\sqrt{t}}$

$\lambda _{D}=\frac{h}{mv}$

$\therefore\; \; v=at$

$v=\frac{eE}{m}t\; \; \; \; \left ( a=\frac{eE}{m} \right )$

$\lambda _{D}=\frac{h}{m\left ( \frac{eE}{m}t \right )}$

$\lambda _{D}=\frac{h}{eEt}$

$\frac{d\lambda _{d}}{dt}=-\frac{h}{\left | e \right |Et^{2}}$

Hence the correct option is (1).

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Posted by

avinash.dongre

JEE Main high-scoring chapters and topics

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