Science (English Medium) कक्षा ११ - CBSE Question Bank Solutions for Chemistry

Dual behaviour of matter proposed by de Broglie led to the discovery of electron microscope often used for the highly magnified images of biological molecules and other type of material. If the velocity of the electron in this microscope is 1.6 × 10⁶ ms^–1, calculate de Broglie wavelength associated with this electron.

Similar to electron diffraction, neutron diffraction microscope is also used for the determination of the structure of molecules. If the wavelength used here is 800 pm, calculate the characteristic velocity associated with the neutron.

The velocity associated with a proton moving in a potential difference of 1000 V is 4.37 × 10⁵ ms^–1. If the hockey ball of mass 0.1 kg is moving with this velocity, calculate the wavelength associated with this velocity.

The critical temperature  for carbon dioxide and methane are 31.1 °C and –81.9 °C respectively. Which of these has stronger intermolecular forces and why?

ΔU^⊖of combustion of methane is – X kJ mol^–1. The value of ΔH^⊖ is ______.

The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are, - 890.3 kJ mol^-1 -393.5 kJ mol^-1, and - 285.8 kJ mol^-1 respectively. Enthalpy of formation of CH_4(g) will be ______.

Explain why alkyl groups act as electron donors when attached to a π system

Write chemical reactions to justify that hydrogen peroxide can function as an oxidizing as well as reducing agent.

According to de Broglie, matter should exhibit dual behaviour, that is both particle and wave like properties. However, a cricket ball of mass 100 g does not move like a wave when it is thrown by a bowler at a speed of 100 km/h. Calculate the wavelength of the ball and explain why it does not show wave nature.

Assertion (A): All isotopes of a given element show the same type of chemical behaviour.

Reason (R): The chemical properties of an atom are controlled by the number of electrons in the atom.

Gases possess characteristic critical temperature which depends upon the magnitude of intermolecular forces between the particles. Following are the critical temperatures of some gases.

From the above data what would be the order of liquefaction of these gases? Start writing the order from the gas liquefying first

Two different gases ‘A’ and ‘B’ are filled in separate containers of equal capacity under the same conditions of temperature and pressure. On increasing the pressure slightly the gas ‘A’ liquefies but gas B does not liquify even on applying high pressure until it is cooled. Explain this phenomenon.

The critical temperature (T_c) and critical pressure (p_c) of \[\ce{CO2}\] are 30.98°C and 73 atm respectively. Can \[\ce{CO2 (g)}\] be liquefied at 32°C and 80 atm pressure?

Assertion (A): Gases do not liquefy above their critical temperature, even on applying high pressure.

Reason (R): Above critical temperature, the molecular speed is high and intermolecular attractions cannot hold the molecules together because they escape because of high speed.

Assertion (A): At critical temperature liquid passes into gaseous state imperceptibly and continuously.

Reason (R): The density of liquid and gaseous phase is equal to critical temperature.

Why does the boundary between liquid phase and gaseous phase disappear on heating a liquid upto critical temperature in a closed vessel? In this situation what will be the state of the substance?

Assertion (A): Combustion of all organic compounds is an exothermic reaction.

Reason (R): The enthalpies of all elements in their standard state are zero.

In which of the following reactions, the equilibrium remains unaffected on addition of small amount of argon at constant volume?

Assertion (A): In the dissociation of \[\ce{PCl5}\] at constant pressure and temperature addition of helium at equilibrium increases the dissociation of \[\ce{PCl5}\].

Reason (R): Helium removes \[\ce{Cl2}\] from the field of action.

Calculate the strength of 5 volume \[\ce{H2O2}\] solution.