Tag: chemistry

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Questions Related to chemistry

Multiple choice chemistry atom fundamental particles of an atom the structure of atoms discovery of subatomic particles

The total number of electrons present in 1.8 g of water is :

  1. $6.023\times 10^{22}$
  2. $10.8576\times 10^{23}$
  3. $10.8576\times 10^{22}$
  4. $6.023\times 10^{23}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

We know that, 1 molecule of water consists of 10 electrons (2 electrons each of two hydrogen atoms and 8 electrons of one oxygen atom).
The number of molecules present in 1.8 g of water.
$18 : g : of : H _2O\xrightarrow[]{contains}6.023\times 10^{23}molecules$
$\Rightarrow1.8 : g : of : H _2O : contains : '! x' : molecules$
$x=\dfrac{1.8\times 6.023\times 10^{23}}{18}$
$=\dfrac{ 6.023\times 10^{23}}{10}=6.023\times 10^{22}molecules$.
$\therefore$ Number of electrons in 1.8 g of water $=6.023\times 10^{22}\times10=6.023\times 10^{23}$

Multiple choice chemistry atom fundamental particles of an atom the structure of atoms discovery of subatomic particles

How many moles of electron weighs one kilogram?

  1. $6.023\times 10^{23}$
  2. $\dfrac {1}{9.108}\times 10^{31}$
  3. $\dfrac {6.023}{9.108}\times 10^{54}$
  4. $\dfrac {1}{9.108\times 6.023}\times 10^8$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Mass of one electron is $9.108\times 10^{-31} kg$.


Mass of one mole of electrons is ${9.108 \times 10^{-31} \times 6.023}\times 10^{23}={9.108\times 6.023}\times 10^{-8}$.

Thus, $1$ kg corresponds to $\dfrac {1}{9.108\times 6.023}\times 10^8$ moles of electrons.

Hence, the correct option is $D$

Multiple choice chemistry atom fundamental particles of an atom the structure of atoms discovery of subatomic particles

How many moles of electrons weigh one kilogram? 

(Mass of electron = $\displaystyle 9.108\times 10^{-31}kg $; Avagadro number = $\displaystyle 6.023\times 10^{23}kg $)

  1. $\displaystyle \frac{1}{9.108\times 6.023}\times 10^{8} $
  2. $\displaystyle 6.023\times 10^{23}$
  3. $\displaystyle \frac{1}{9.108}\times 10^{31}$
  4. $\displaystyle \frac{6.023}{9.108}\times 10^{54}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Mass of electron $\displaystyle 9.108\times 10^{-31}kg$.
The number of electrons that weigh 1 kg will be $\frac {1}{\displaystyle 9.108\times 10^{-31}}$
The Avagadro number is $\displaystyle 6.023\times 10^{23}$.    
The number of moles of electrons that weigh 1 kg will be
$\frac {1}{\displaystyle 9.108\times 10^{-31} \times \displaystyle 6.023\times 10^{23}}=\displaystyle \frac{1}{9.108\times 6.023}\times 10^{8}$

Multiple choice chemistry acids and alkalis neutralisations in everyday life acids and bases in daily life neutralisation reaction

Basic soil can be corrected by adding:

  1. urea

  2. nitric acid

  3. organic manure

  4. formic acid
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Plants can neither grow in too acidic conditions nor grow in too basic conditions. Thus for the proper growth of the plant optimum pH of the soil has to be maintained.


If the soil is too basic then it can be corrected by using organic manure. Compost micro-organisms operate best under neutral to acidic conditions with their pH ranging from 5.8 to 8. 


Thus they not only help in maintaining the pH of the soil but also increase the nutrient availability and improve the quality of the soil (unlike inorganic chemical fertilizers which damage the soil quality). 

Thus organic manure creates acidic conditions, which help in neutralizing the basic soil and makes the pH optimum for plant growth.

Hence, the correct answer is the option (C).

Multiple choice chemistry acids and alkalis neutralisations in everyday life acids and bases in daily life neutralisation reaction

A 26 ml of $ N-Na _{2}CO _{3} $ solution is neutralized by the solutions of acids A and B in different experiments. The volumes of the acids A and B required were $10 ml$ and $40 ml$, respectively. How many volumes of A and B are to be mixed in order to prepare 1 litre of normal acid solution? 

  1. $179.4, 820.6$
  2. $820.6, 179.4$
  3. $500, 500$
  4. $474.3, 525.7$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$ N _{1}V _{1} (Na _{2}CO _{3}) = N _{2}V _{2} (A)$
$ N _{1}V _{1} (Na _{2}CO _{3}) = N _{2}V _{2}(B)$
$ N _{1} = 1 , V _{1} = 26ml, V _{2} = 10ml, V _{3} = 40ml$
Normality of A
$ N _{2} = \dfrac{N _{1}V _{1}}{V _{2}} = \dfrac{1 \times 26}{10} = 2.6 N $
Normality of B
$ N _{3}= \dfrac{N _{1}V _{1}}{V _{3}} = \dfrac{1 \times 26}{40} = 0.65 N $
if we mix A & B then
$ N _{1}V _{1} + N _{2} V _{2} = N _{3} (V _{1}+V _{2})$
$ 2.6 V _{1}+0.65V _{2} = 1 \times 1000 $
$ 2.6V _{1}+ V _{2} + 0.65 V _{2} = 1000 ...(1)$
$ V _{1}+V _{2} =1000 ...(2)$
multiply $eq^{n}$ (2) by 0.65 
$ 0.65 V _{1}+0.65V _{2} = 650 ...(3)$
$ V _{2} = 820.6 ml$
$ V _{1} = 1000 - 820.6 = 179.4 ml $
option "a" correct.

Multiple choice chemistry acids and alkalis neutralisations in everyday life acids and bases in daily life neutralisation reaction

What volume of $ 0.18 N - KMnO _{4} $ solution would be needed for complete reaction with 25 ml of $ 0.21 N - KNO _{2} $ in acidic medium ?

  1. $57.29 ml$
  2. $11.67 ml$
  3. $29.17 ml$
  4. $22.92 ml$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$ KMnO _{4}+KNO _{2}$
$ 0.18 N $ 25 ml
V = ? 0.21 N
we know $ N _{1}V _{1}= N _{2}V _{2}$
$ 0.18 \times V _{1} = 0.21 \times 25 $
$ V _{1} = \dfrac{0.21 \times 25}{0.18 }$
$ \boxed{V _{1} = 29.17 ml}$
Answer option C

Multiple choice chemistry acids and alkalis neutralisations in everyday life acids and bases in daily life neutralisation reaction

Which one of the following types of medicines is for treating indigestion ? 

  1. Antibiotic

  2. Analgesic

  3. Antacid

  4. Antiseptic
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Indigestion can lead to a feeling of fullness, burping, heartburn (due to gastric acid reflux), etc. Thus an antacid can neutralize the effect of an excess of acid and hence it can be consumed for some relief from heartburn, indigestion, acidity, etc.

Antibiotics are taken in case of bacterial infections, Analgesic act as painkillers, Antiseptic is to prevent sepsis.
Hence, the correct answer is the option (C).

Multiple choice chemistry introduction to acid and base neutralisations in everyday life acids and bases in daily life neutralisation reaction

Calcium hydroxide (slaked lime) is used in :

  1. plastics and dyes

  2. fertilizers

  3. antacids

  4. white washing
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Calcium hydroxide called slaked lime is an inorganic compound. It is a colorless crystal and is obtained when calcium oxide is mixed with water. It is used in preparation of dry mixes for painting and white washing.

Multiple choice chemistry introduction to acid and base neutralisations in everyday life acids and bases in daily life neutralisation reaction

Caustic soda is the common name for:

  1. $\displaystyle { Mg\left( OH \right) } _{ 2 }$
  2. $\displaystyle KOH$
  3. $\displaystyle { Ca\left( OH \right) } _{ 2 }$
  4. $\displaystyle NaOH$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Caustic soda is sodium hydroxide which is a strong base and so it has high basicity value on $pH$ scale. It is widely used in acid-base titrations to neutralize various strong or weak acids like sulfuric acid.


Hence, the correct option is $\text{D}$