Tag: experiments with water

Questions Related to experiments with water

 What happens on adding salt to the water in which eggs are boiled?  

  1. Egg will take more time to cook

  2. Egg will start floating on water

  3. Egg will cook faster

  4. Egg will sink in water


Correct Option: C

Water is liquid at room temperature, the most important reason for this is the :

  1. high boiling point of water

  2. high melting point of water

  3. high heat of vapourization of water

  4. cohesive forces due to hydrogen bonds in water


Correct Option: D
Explanation:

Water is liquid at room temperature, the most important reason for this is the cohesive forces due to hydrogen bonds in water.

Which of the following is true?

  1. Vegetable oil is heavier than water

  2. Ghee is soluble in water

  3. Curd is soluble in water

  4. A person who can not swim will not drown in dead sea


Correct Option: D
Explanation:

The salt is all dissolved in the water, this makes all of the water much denser than normal water, making it uncomfortable to swim in. Furthermore it means that the water is denser than the human body which is 70%, unsalted water. Meaning it is impossible to sink in the Dead Sea.

One litre of a samle of hard water contains 1 mg of $CaCl _2$ and 1 mg of $MgCl _2$.Then the total hardness in terms of parts of $CaCO _3$ per $10^8$ parts of water by mass is:

  1. 1.954 ppm

  2. 1.260

  3. 0.946

  4. none of these


Correct Option: A
Explanation:

$1$ mole of $CaCl _2\equiv1$ mole of $CaCO _3\equiv1$ mole of $MgCl _2$

$\therefore$ $100mg$ $CaCO _3$ is produced by $95mg$ $MgCl _2$
$\therefore$ $1mg$ of $MgCl _2$ gives $\cfrac {100}{95}mg$  $CaCO _3=$ $1.05mg$ $CaCO _3$

Similarly, $1mg$ of $CaCl _2$ gives $\cfrac {100}{111}mg$ $CaCO _3=$ $0.90mg$ $CaCO _3$

$\therefore$ Total $CaCO _3$ per litre of water= $1.05+0.90=1.95$ $mg$
Weight of $1000ml$ of water= $10^3g$=$10^6mg$

$\therefore$ Total hardness in terms of parts of $CaCO _3$ per $10^6$ parts of water by weight = $\cfrac {1.95}{10^6}\times 10^6$
=$1.95$ $ppm$

Answer: (A) $1.954$ $ppm$

100 ml of tap water containing $Ca(HCO _3) _2$ was titrated with N/50 $HCl$ with methyl orange as indicator. If 30ml of $HCl$ were required, calculate the temporary hardness as part of $CaCO _3$ per $10^6$ parts of water.

  1. 150 ppm

  2. 300 ppm

  3. 450 ppm

  4. 600 ppm


Correct Option: B
Explanation:

Milli equivalent of $Ca(HCO _{3}) _{2}$= miili equivalent of $CaCO _{3}$= mili equivalent of $HCl$

$N _{1}V _{1}= N _{2}V _{2}$
$1000 \times w$/($100$/$2$)= $30$/$50$
$w$= $0.03$ g
For $100$ml the amount of $CaCO _{3} = 0.03$ g
For $10^{6}$ = $0.03 \times 10^{6}$/$100=300$ ppm

The density of water is less in its solid state because:

  1. in solid state(ice) , water molecules are arranged in highly ordered open cage like structure

  2. more extensive hydrogen bonding is present in solid state ice

  3. the water molecules are closet in solid state of water

  4. water is a rigid crystalline, closely packed structure in its solid state.


Correct Option: A
Explanation:

The water has maximum density at 4 deg C. When liquid water is converted to ice, an open cage three dimensional structure is formed. Almost half the space is unoccupied. In liquid state, molecules are more closely packed.

The formula of calgon is:

  1. $Na _{2}\left [ Na _{4}\left ( PO _{3} \right ) _{6} \right ]$

  2. $Na _{4}\left [ Na _{2}\left ( PO _{3} \right ) _{6} \right ]$

  3. $Na _{4}\left [ Na _{2}\left ( PO _{3} \right ) _{3} \right ]$

  4. $Na _{2}\left [ Na _{4}\left ( PO _{3} \right ) _{4} \right ]$


Correct Option: A
Explanation:
The formula of calgon is $Na _2[Na _4(PO _3) _6]$. Calgon is a trade name of a complex salt, sodium hexametaphosphate $(NaPO _3) _6$. It is used for softening hard water. 
Calgon ionizes to give a complex anion.

The process used for the removal of hardness of water is:

  1. Baeyer

  2. Calgon

  3. Hoope

  4. Serpeck


Correct Option: B
Explanation:

Hard water is  percolated over Calgon (sodium hexameta phosphate $Na _6P _6O _{18}$). Calgon reacts calcium ( or magnesium) ions present in hard water. Inactive complex ions are formed that do not precipitate with soap.

Which of the following is not a disadvantage of using hard water?

  1. In production of steam in boilers

  2. Formation of scales in cooking utensils

  3. In cooking,bathing and washing

  4. In ion exchanges


Correct Option: D
Explanation:

In an ion exchange using hard water is not a disadvantage since it becomes soft by exchanging anions or cations. $Na^+$, $Ca^{2+}$ and $Mg^{2+}$ ions (from water) are exchanged with $H^+$ ions (from cation exchange resins). $Cl^-$, $SO _4^{2-}$ and $NO _3^{-}$ ions (from water) are exchanged with $OH^-$ ions (from anion exchange resins).

Polyphosphates like sodium hexametaphosphate (calgon) are used as water softening agents because they:

  1. form soluble complexes with anionic species

  2. precipitate anionic species

  3. form soluble complexes with cationic species

  4. precipitate catatonic species.


Correct Option: C
Explanation:

 Polyphosphates like sodium hexametaphosphate (calgon) form souble complex with cations like $Ca^{2+}$ and  $Mg^{2+}  $present in hard water and hence are used as water softening agents.
$Na _{6}P _{6}O _{18} \rightarrow 2Na^{+}+ Na _{4}P _{6}O^{2-} _{18}$
$M^{2+} + Na _{4}P _{6}O^{2-} _{18}\rightarrow [Na _{2}MP _{6}O _{18}]^{2-} + 2Na^{+}$