Tag: relative formula mass

Questions Related to relative formula mass

The charge to mass ratio of $\alpha - particle$ is approximately ____  of the charge to mass ratio of a proton is:

chemistry quantitative chemistry molecular mass relative formula mass masses of atoms and molecules

  1. half

  2. twice

  3. 4 times

  4. 6 times

Show answer
Correct Option: B
Explanation:
Alpha particle is a helium nucleus containing two protons and two neutrons so its charge is $twice$ the proton’s charge while the mass is about 4 times greater.
Let the charge of the proton be $+e$, then the charge of the alpha particle will be $+2e$. Similarly mass of the proton be m, then the mass of the alpha particle will be $4m$.
Specific charge = charge/mass of the substance.
For proton, specific charge =$\dfrac { e }{ m } $
For alpha particle, specific charge = $\dfrac { 2e }{ 4m } $
Therefore the ratio is = $\dfrac { e }{ m } $ x $\dfrac { 4m }{ 2e }  = 2:1$
Hence, the charge to mass ratio of an alpha particle is approximate twice the charge to mass ratio of a proton.  
So, the correct option is $B$

How many grams of phosphoric acid would be needed to neutralize $100$ gm of magnesium hydroxide? (Molecular weight of $H _3PO _4=98$ and $Mg(OH) _2=58.3 gm$)

chemistry quantitative chemistry molecular mass relative formula mass masses of atoms and molecules

  1. 66.7 gm

  2. 252 gm

  3. 112 gm

  4. 168 gm

Show answer
Correct Option: C

The weight of $1$ litre of a glass at STP is $2$ grams, its molecular weight is:

chemistry quantitative chemistry molecular mass relative formula mass masses of atoms and molecules

  1. $44.4$

  2. $44.8$

  3. $44.1$

  4. $55.8$

Show answer
Correct Option: B
Explanation:

$At\quad STP,\quad volume\quad of\quad 1\quad mole\quad gas\quad =\quad 22.4\quad L\ Mass\quad of\quad 1L\quad gas\quad =\quad 2g\ Mass\quad of\quad 22.4L\quad gas\quad =\quad 2\times 22.4\quad =\quad 44.8\ So,\quad molecular\quad weight\quad of\quad gas\quad =\quad 44.8\ So,\quad correct\quad answer\quad is\quad option\quad B.$

The number of gram-molecules of oxygen in $6.022\times 10^{24}$ molecules of $CO$ is:

chemistry quantitative chemistry molecular mass relative formula mass masses of atoms and molecules

  1. $10$ gm moles

  2. $5$ gm moles

  3. $1$ gm mole

  4. $0.5$ gm mole

Show answer
Correct Option: B
Explanation:

Number of oxygen atoms = Number of $CO$ molecules$=6.022\times 10^{24}$

Number of oxygen molecule $=\dfrac12\times$  Number of oxygen atoms $=3.011\times 10^{24}$
Number of g-molecule of $O _2$ molecules $=\dfrac{3.011\times 10^{24}}{6.022\times 10^{23}}=5\ gm\ mole$

The molar mass of $CuSO _4.5H _2O$ is 249. Its equivalent mass in the reaction (a) and (b) would be:
(a) Reaction $CuSO _4 + KI \rightarrow$ product
(b) Electrolysis of $CuSO _4$ solution

chemistry quantitative chemistry molecular mass relative formula mass masses of atoms and molecules

  1. (a) 249 (b) 249

  2. (a) 124.5 (b) 124.5

  3. (a) 249 (b) 124.5

  4. (a) 124.5 (b) 249

Show answer
Correct Option: B
Explanation:

In water, ${ CuSO } _{ 4 }\cdot 5{ H } _{ 2 }O$ dissociates into ${ Cu }^{ 2+ }$ & ${ SO } _{ 4 }^{ 2- }$. The ions are doubly charged. Hence half as much is needed to react with a compound that dissociates into singly charged species.

(a) Reaction of ${ CuSO } _{ 4 }+KI\longrightarrow $ products.
     Eq. wt of ${ CuSO } _{ 4 }=249/2=124.5$
(b) Electrolysis of ${ CuSO } _{ 4 }$
     At Cathode ${ \underset { +2 }{ Cu }  }^{ 2+ }+{ 2e }^{ - }\longrightarrow \underset { 0 }{ Cu } $
$\therefore$   Charge in oxidation state $=2=n-$factor
$\therefore$   Eq. wt $=\dfrac { 249 }{ 2 } =124.5$

10 ml of 0.1 M solution sodium hydroxide is completely neutralised by 25 ml of 3 gram of dibasic acid in one solution the molecular weight of acid is:

chemistry quantitative chemistry molecular mass relative formula mass masses of atoms and molecules

  1. 225 g

  2. 250 g

  3. 300 g

  4. 150 g

Show answer
Correct Option: D
Explanation:
Given,
Concentration of $NaOH=0.1 N$
Volume$=10\,mL$
Mass of Dibasic acid$=3g$
Volume$=25\,mL$
Using,
$n _{1}M _{1}V _{1}=n _{2}M _{2}V _{2}$
$n _{2}=2$
$\Rightarrow 1\times 0.1\times 10=2\times25\times\cfrac{3}{M}$
$\Rightarrow M=\cfrac{2\times 25\times 3}{10\times 0.1}$
$\Rightarrow M=150\,g$

M g of a substance when vaporised occupy a volume of 5.6 litre at NTP. The molecular mass of the substance will be: 

chemistry quantitative chemistry molecular mass relative formula mass masses of atoms and molecules

  1. $M$

  2. $2M$

  3. $3M$

  4. $4M$

Show answer
Correct Option: D
Explanation:
Given,
$Mg$ of substance occupy volume$=5.6\,litre $ at NTP.
At NTP,
1 mol occupy 22.4 litre of volume.
5.6 litre$=Mg$
22.4 litres$=4\,Mg$ of substance.
So, Molecular mass of gas$=4\,Mg/mol$