Answer:
V=5.36L
Explanation:
Given:
m=32 g [tex]NO_{2}[/tex]
P=3.12 atm
T=18°C+273.15=291.15 K
R=0.08205 [tex]\frac{L*atm}{mol*K}[/tex] [gas constant]
[note temperature should always be in absolute value (kelvin) thus we convert temperature by using K= °C+273.15]
Required:
Volume of [tex]NO_{2}[/tex]
Solution:
Based on the given Ideal Gas Law will be used to solve for the volume but first solve for the moles present by multiplying the given mass to the molar mass of the compound
N: 1 × 14 [tex]\frac{g}{mol}[/tex]= 14 [tex]\frac{g}{mol}[/tex]
O: 2 × 16[tex]\frac{g}{mol}[/tex]= 32 [tex]\frac{g}{mol}[/tex]
Molar mass: 14 [tex]\frac{g}{mol}[/tex] + 32[tex]\frac{g}{mol}[/tex]= 46 [tex]\frac{g}{mol}[/tex]
moles of [tex]NO_{2}[/tex]= 32g [tex]NO_{2}[/tex] ×[tex]\frac{1 mol NO_{2}}{46g NO_{2}}[/tex]=0.7mol [tex]NO_{2}[/tex]
~Using the Ideal Gas formula~
[tex]PV=nRT[/tex] ⇒ [tex]V=\frac{nRT}{P}[/tex]
[tex]V=\frac{(0.7mol)(0.08205\frac{L*atm}{mol*K})(291.15K) }{3.12 atm}[/tex]
V=5.36L
