Hydrogen chloride gas (7.3 g) is in a container at a pressure of 2.5 atm and a temperature of 300 K.
(1.) How many moles of gas are present?
(2.) How many molecules of gas are present?
(3.) If all the gas were dissolved in 100 cm³ of water, what mass of silver nitrate would need to be added to precipitate all the chloride ion produced.
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QUESTION:-
Hydrogen chloride gas (7.3 g) is in a container at a pressure of 2.5 atm and a temperature of 300 K.
(1.) How many moles of gas are present?
(2.) How many molecules of gas are present?
(3.) If all the gas were dissolved in 100 cm³ of water, what mass of silver nitrate would need to be added to precipitate all the chloride ion produced.
[tex]\rule{100}{3}\bigstar\bigstar\bigstar\rule{100}{3}[/tex]
EXPLANATION:-
(i) How many moles of gas is present ?
We know that-;
[tex]\longrightarrow \boxed{\bigstar\bf\;Moles=\frac{mass}{molar\;mass}\bigstar}[/tex]
In the given question-:
Moles =??
Mass=7.3 g (GIVEN)
Molar mass= 36.46 g/mol
Putting value in the formula we get -:
[tex]\rightarrow\ \bf\;Moles=\frac{3.6}{36.46}\\\\\\\rightarrow \bf\;Moles=\frac{730}{3646} \; \\\\\ \longrightarrow \underline{\bf\;0.20\;Mol}[/tex] [Kindly refer 1st attachment for the divison]
[tex]\rule{100}{3}\bigstar\bigstar\bigstar\rule{100}{3}[/tex]
(ii)How many molecules of gas are present?
We know that-:
[tex]\longrightarrow \boxed{\bigstar\bf\;Molecules=Moles\times Avogadro's\; constant(6.022\times10^{23)}\bigstar}[/tex]
Molecules= ??
Moles-0.20
Arogardo's constant=6.022×10²³
Putting value-:
[tex]\rightarrow \bf\;Molecules=0.20\times6.022\times10^{23}\\\\\\\rightarrow \bf\;Molecules=1.2044\times10^{23}\\\\\\\longrightarrow \underline{\bf\;Molecules=1.204\times10^{23}}[/tex]
[tex]\rule{100}{3}\bigstar\bigstar\bigstar\rule{100}{3}[/tex]
(3) If all the gas were dissolved in 100 cm³ of water, what mass of silver nitrate would need to be added to precipitate all the chloride ion produced.
To find this first we will balance the equation of the reaction -:
AgNO₃+HCL→AgCl+HNO₃
Observing reaction we get-:
Moles of AgNO₃=Moles of HCL
Moles of AgNO₃=0.20 Mol (Calculated above)
Now-:
[tex]\longrightarrow \boxed{\bigstar\bf\;Moles=\frac{mass}{molar\;mass}\bigstar}[/tex]
Moles=0.20
Mass=??
Molar mass= 169.91
Putting values we get-:
[tex]\rightarrow \bf\;0.20=\frac{mass}{169.91}\\\\\\\rightarrow \bf\;Mass=169.91\times0.20\\\\\\\longrightarrow \underline{\bf\;Mass=33.982\;g}\\\\\\[/tex]
Therefore of mass=33.982 or mass≈34 g
[REFER 2ND ATTACHMENT FOR MULTIPLICATION]
[tex]\rule{100}{3}\bigstar\bigstar\bigstar\rule{100}{3}[/tex]
Explanation:
QUESTION:-
Hydrogen chloride gas (7.3 g) is in a container at a pressure of 2.5 atm and a temperature of 300 K.
(1.) How many moles of gas are present?
(2.) How many molecules of gas are present?
(3.) If all the gas were dissolved in 100 cm³ of water, what mass of silver nitrate would need to be added to precipitate all the chloride ion produced.
EXPLANATION:-
(i) How many moles of gas is present ?
We know that-;
\longrightarrow \boxed{\bigstar\bf\;Moles=\frac{mass}{molar\;mass}\bigstar}⟶
★Moles=
molarmass
mass
★
In the given question-:
Moles =??
Mass=7.3 g (GIVEN)
Molar mass= 36.46 g/mol
Putting value in the formula we get -:
\begin{gathered}\rightarrow\ \bf\;Moles=\frac{3.6}{36.46}\\\\\\\rightarrow \bf\;Moles=\frac{730}{3646} \; \\\\\ \longrightarrow \underline{\bf\;0.20\;Mol}\end{gathered}
→ Moles=
36.46
3.6
→Moles=
3646
730
⟶
0.20Mol
[Kindly refer 1st attachment for the divison]
\rule{100}{3}\bigstar\bigstar\bigstar\rule{100}{3}
(ii)How many molecules of gas are present?
We know that-:
\longrightarrow \boxed{\bigstar\bf\;Molecules=Moles\times Avogadro's\; constant(6.022\times10^{23)}\bigstar}⟶
★Molecules=Moles×Avogadro
′
sconstant(6.022×10
23)
★
Molecules= ??
Moles-0.20
Arogardo's constant=6.022×10²³
Putting value-:
\begin{gathered}\rightarrow \bf\;Molecules=0.20\times6.022\times10^{23}\\\\\\\rightarrow \bf\;Molecules=1.2044\times10^{23}\\\\\\\longrightarrow \underline{\bf\;Molecules=1.204\times10^{23}}\end{gathered}
→Molecules=0.20×6.022×10
23
→Molecules=1.2044×10
23
⟶
Molecules=1.204×10
23
\rule{100}{3}\bigstar\bigstar\bigstar\rule{100}{3}
(3) If all the gas were dissolved in 100 cm³ of water, what mass of silver nitrate would need to be added to precipitate all the chloride ion produced.
To find this first we will balance the equation of the reaction -:
AgNO₃+HCL→AgCl+HNO₃
Observing reaction we get-:
Moles of AgNO₃=Moles of HCL
Moles of AgNO₃=0.20 Mol (Calculated above)
Now-:
\longrightarrow \boxed{\bigstar\bf\;Moles=\frac{mass}{molar\;mass}\bigstar}⟶
★Moles=
molarmass
mass
★
Moles=0.20
Mass=??
Molar mass= 169.91
Putting values we get-:
0.20=mass/169.91
mass=0.20×169.91
mass=34 g
\rule{100}{3}\bigstar\bigstar\bigstar\rule{100}{3}