The subscript 2 indicates number of Cl atoms
<h3>Further explanation
</h3>
The molecular formula shows the number of atoms of a compound
whereas the empirical formula is the simplest comparison of the atoms making up the compound
MgCl₂ is an ionic compound composed of Mg and Cl atoms
Mg is in group 2 which has a +2 charge, while Cl is in the Halogen group which has a charge of -1
If the two are bonded, the 2 electrons released by Mg will be bound by 2 Cl atoms, each of which atoms of Cl need 1 more electron in order to obtain a stable electron configuration such as a noble gas
So the number 2 in the MgCl compound indicates the number of Cl atoms
Explanation:
Mass of solution , m= 0.94 g + 215 g = 215.94 g
Volume of the solution = V
Density of solution = d = ![10.8 g/cm^3](https://tex.z-dn.net/?f=10.8%20g%2Fcm%5E3)
![V=\frac{m}{d}=\frac{215.94 g}{10.8 g/cm^3}=19.99 cm^3=19.99 mL](https://tex.z-dn.net/?f=V%3D%5Cfrac%7Bm%7D%7Bd%7D%3D%5Cfrac%7B215.94%20g%7D%7B10.8%20g%2Fcm%5E3%7D%3D19.99%20cm%5E3%3D19.99%20mL)
![1 cm^3=1 mL](https://tex.z-dn.net/?f=1%20cm%5E3%3D1%20mL)
![1 mL=0.001 L](https://tex.z-dn.net/?f=1%20mL%3D0.001%20L)
![V=0.01999 L](https://tex.z-dn.net/?f=V%3D0.01999%20L)
Moles of hydrogen = ![\frac{0.94 g}{2 g/mol}=0.47 mol](https://tex.z-dn.net/?f=%5Cfrac%7B0.94%20g%7D%7B2%20g%2Fmol%7D%3D0.47%20mol)
Molarity of hydrogen gas:
![Molarity=\frac{Moles}{\text{Volume of solution (L)}}](https://tex.z-dn.net/?f=Molarity%3D%5Cfrac%7BMoles%7D%7B%5Ctext%7BVolume%20of%20solution%20%28L%29%7D%7D)
![=\frac{0.47 mol}{0.01999 L}=0.02351 mol/L](https://tex.z-dn.net/?f=%3D%5Cfrac%7B0.47%20mol%7D%7B0.01999%20L%7D%3D0.02351%20mol%2FL)
Molality of hydrogen gas :
Mass of solvent or here it is palladium = 215 g =0.215 kg (1 g = 0.001 kg)
![Molality=\frac{Moles}{\text{Mass of solvent(kg)}}](https://tex.z-dn.net/?f=Molality%3D%5Cfrac%7BMoles%7D%7B%5Ctext%7BMass%20of%20solvent%28kg%29%7D%7D)
![=\frac{0.47}{0.215 kg}=2.1860 mol/kg](https://tex.z-dn.net/?f=%3D%5Cfrac%7B0.47%7D%7B0.215%20kg%7D%3D2.1860%20mol%2Fkg)
Mass percentage of hydrogen is solution :
![\frac{\text{Mass of hydrogen}}{\text{Mass of solution}}\times 100](https://tex.z-dn.net/?f=%5Cfrac%7B%5Ctext%7BMass%20of%20hydrogen%7D%7D%7B%5Ctext%7BMass%20of%20solution%7D%7D%5Ctimes%20100)
![=\frac{0.94 g}{215.94 g}\times 100=0.435\%](https://tex.z-dn.net/?f=%3D%5Cfrac%7B0.94%20g%7D%7B215.94%20g%7D%5Ctimes%20100%3D0.435%5C%25)
The percent yield : 73.5%
<h3>Further explanation</h3>
Given
Reaction
C+2H₂⇒CH₄
Required
The percent yield
Solution
mol of Carbon(as a limiting reactant) :
![\tt \dfrac{100}{12}=8.3](https://tex.z-dn.net/?f=%5Ctt%20%5Cdfrac%7B100%7D%7B12%7D%3D8.3)
mol CH₄ based on C, and from equation mol ratio C : CH₄, so mol CH₄ = 8.3
Mass of Methane(theoretical yield) :
![\tt mass=mol\times MW\\\\mass=8.3\times 16=133.3~g](https://tex.z-dn.net/?f=%5Ctt%20mass%3Dmol%5Ctimes%20MW%5C%5C%5C%5Cmass%3D8.3%5Ctimes%2016%3D133.3~g)
![\tt \%~yield=\dfrac{actual}{theoretical}\times 100\%\\\\\%yield=\dfrac{98}{133.3}\times 100\%=73.5\%](https://tex.z-dn.net/?f=%5Ctt%20%5C%25~yield%3D%5Cdfrac%7Bactual%7D%7Btheoretical%7D%5Ctimes%20100%5C%25%5C%5C%5C%5C%5C%25yield%3D%5Cdfrac%7B98%7D%7B133.3%7D%5Ctimes%20100%5C%25%3D73.5%5C%25)