# Haryana Board Class 12 Chemistry Questions with Answer Exam 2021

**Very Short Answer Type Questions: 2Marks Questions**

## A compound is formed by two elements M and N. The element N forms CCP and atoms of M occupy 1/3rd of tetrahedral voids. What is the formula of the compound ?

Answer:

The ccp lattice is formed by the atoms of the element N.

We know that-

the number of tetrahedral voids = twice the number of atoms of the element N.

The atoms of element M occupy 1/3rd of the tetrahedral voids.

Therefore, the number of atoms of M is equal to 2 x 1/3 = 2/3rd of the number of atoms of N.

Therefore, ratio of the number of atoms of M to that of N is M : N = (2/3):1 = 2:3.

Thus, the formula of the compound is M_{2}N_{3}

## Concentrated nitric acid used in laboratory work is 68% nitric acid by mass in aqueous solution. What should be the molarity of such a sample of the acid if the density of the solution is 1.504 g mL^{-1} ?

Answer:

68% of nitric acid by mass = Mass of nitric acid = 68gm

Mass of solution = 100g

Molar mass of nitric acid = 63g mol^{-1}

Hence, 68gm nitric acid = 1.079 mole

Density of solution = 1.504g mL^{-1}

So, Volume of solution = 100/105.4 = 66.5 ml. = 0.0665L

now, molarity of the solution = Mass of solute/Volume of solution in liter

= 1.079/0.0665 = 16.2mol/L = 16.2M

## Calculate the mass of a non-volatile solute (molar mass 40 g mol^{-1}) which should be dissolved in 114 g octane to reduce its vapour pressure to 80%.

Answer:

Let the vapour pressure of pure octane be p^{o}

Then, the vapour pressure of the octane after dissolving the non-volatile solute is (80/100)p^{o} = 0.8 p^{o}

Molar mass of solute (M_{2}) = 40 gm mol^{-1}

Mass of octane (w_{1}) = 114 gm

Molar mass of octane (M_{1}) = 8 × 12 + 18 × 1 = 114 g mol^{-1}

According to Raoult's law,

(p^{o} - p) / p^{o} = (w_{2} x M_{1} ) / (M_{2} x w_{1})

⇒ (p^{o} - 0.8 p^{o}) / p^{o} = (w_{2} x 114 ) / (40 x 114)

⇒ 0.2 p^{o} / p^{o} = w_{2} / 40

⇒ 0.2 = w_{2} / 40

⇒ w_{2} = 8 gm

Hence, the required mass of the solute is 8 gm.

## Calculate the half life of a first order reaction from rate constant 500 sec^{−1}

Answer:

Here rate constant k = 500 sec^{−1}

∴ Half – life of a first order reaction is -

t_{1/2} = 0.693/500 = 0.001386 sec. = 1.386 x 10^{−3}Sec.

## What Is Hoffmann bromamide Reaction ?

Answer:

When an amide is treated with bromine in an aqueous or ethanolic solution of sodium hydroxide, degradation of amide takes place leading to the formation of primary amine contains one carbon less than the number of carbon atoms in that amide. This reaction is known as Hoffmann bromamide degradation reaction.

RCONH_{2} +Br_{2} + 4NaOH → R-NH_{2} + Na_{2}CO_{3} + 2NaBr + 2H_{2}O