Relative formula mass - class-IX

The relative atomic mass of  $Ne$ is $20.$ Its relative molecular mass is also $20$ as Neon is mono-atomic.

Weight of $1$ mole of $O _2=32g$

First, calculate the mass of NaCl in 25g of salt (25 * 0.08775 = 2.19375g). Then, divide by the molar mass of NaCl (58.44 g/mol) to get moles, and multiply by Avogadro's number (6.022 * 10^23) to find the formula units.

${S _8} _{(s)}+{8O _2} _{(g)} \longrightarrow {8SO _2} _{(g)}$

Calculate the milliequivalents of ferrous oxalate and ferrous sulphate. Since KMnO4 acts as an oxidizing agent in acidic medium, use the n-factor to determine the volume of 0.1 M solution required to neutralize the total equivalents.

$1g$ atom of $Ag$ implies $1$ mole of it

$S+O _2\rightarrow SO _2$

Number of moles of $Cu$ in $66.6g= \cfrac {66.6 g}{63.5}=1.05$ moles

1 mole of ionic compound contains 1 mole of cations and 2 moles of anions.
The formula mass of the compound is $138+2(19)=176 \ g/mol$.

(A) $ 1 $ mole of any substance contains Avogadro's number of atoms or molecules. Thus, $1$ mole of H atoms corresponds to $=6.02\times 10^{23}H$ atoms.
Hence, the option A is true.
(B) $6.02\times 10^{23}H$ atoms have mass equal to molar mass of H. it is equal to $1.008$ g.
 Hence, the option B is true.
(C) The formula mass of $O _2=32.00 amu$. Thus, $1$ oxygen molecule weighs $32$ amu and $1$ mole of oxygen molecules weighs $32$ g.
Hence, the option C is true.
As we know, 1 mole is the collection of $6.02\times 10^{23}$ entities. Here entities may represent atoms, ions, molecules.
And mass of one mole of H is $1.008$ g so $6.02\times 10^{23}H$ atoms have a mass of $1.008$ g.
The formula mass of $O _2= 2(8p+8n) = 2\times16 = 32.0$ amu.

As we know, one mole of a gaseous substance occupies $22.4$ litres volume.

Given - $ _{17} Cl^{35} - 75.53$% and mass of $Cl$ 34.97 amu
$ _{17} C^{37} - 24.47$% and mass of $Cl$ 36.95 amu
$\therefore $ Atomic weight of chlorine
$= \displaystyle \frac{(75.53 \times 34.97)+(24.47 \times 36.95)}{100} = \frac{3545.45}{100}$
$=35.45$

Valency of $M$ in $MO=2$
Equivalent wt. of $M = 20$
Atomic wt. of $\text{M = Equivalent wt. Valency}$
$;;;;;;;;;;;;;;;;;;;=20\times2=40$

A mole is equal to the relative molecular mass of a gas. As given, the number of moles of gas X and hydrogen are same in the two flasks.
Therefore no. of moles in 2g of hydrogen gas = 1
No. of moles in 1.01g of hydrogen gas = 1/2
Now the weight of gas X that contains 1/2 moles = 16.2g
Weight of gas X that contains 1mole = $\frac{16.2}{1/2}$ = 32g

The formula of the salt with molecular weight 239 is $Ca(ClO _{4}) _{2}$ and thus the divalent metal is calcium with atomic weight 40.

Relative atomic mass or relative molecular mass was used as the mass of the atoms or molecules were very small. Therefore there was a need to be taken relative to some other elements.

Gram atomic mass is the mass, in grams, of one mole of atoms in a monoatomic chemical element. It is numerically equal to the relative atomic mass in grams. The atomic mass of an element expressed in grams is called gram atomic mass.

Given that, 
Mass of $O _2$ = 256 g

 Molecular weight of $NaCl$ is $58.5$ grams,

The balanced reaction is 2H3PO4 + 3Mg(OH)2 -> Mg3(PO4)2 + 6H2O. Calculate the moles of Mg(OH)2 (100/58.3), then use the stoichiometric ratio to find the required moles of H3PO4, and finally convert to grams.

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

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

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.