Answer:
[tex]NO_2[/tex]
Explanation:
Considering,
[tex]n=\frac{m}{M}[/tex]
Using ideal gas equation as:
[tex]PV=\frac{m}{M}RT[/tex]
where,
P is the pressure = 760 mmHg
V is the volume = 100.0 mL = 0.1 L
m is the mass of the gas = 0.193 g
M is the molar mass of the gas = ?
Temperature = 17 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T = (17 + 273.15) K = 290.15 K
R is Gas constant having value = 62.36367 L. mmHg/K. mol
Applying the values as:-
[tex]760\times 0.1=\frac{0.193}{M}\times 62.36367\times 290.15[/tex]
M = 45.95 g/mol
This mass corresponds to [tex]NO_2[/tex]. Hence, the gas must be [tex]NO_2[/tex].
A gas with a molar mass of 47.42 g/mol is likely to be [tex]NO_2[/tex] because the molar mass of [tex]NO_2[/tex] is 46 g/mol.
Given the following data:
Conversion:
Volume = 100-mL to Liters = 0.100 L.
Temperature = 17.0°C to Kelvin = 290 K.
Pressure = 736 mmHg to atm = 0.97 atm.
To determine the unknown gas, we would use the ideal gas law equation;
[tex]PV = \frac{M}{MM} RT[/tex]
Where;
Making MM the subject of formula, we have:
[tex]MM = \frac{MRT}{PV} \\\\MM = \frac{0.193(0.0821)(290)}{0.97(0.100)}\\\\MM = \frac{4.60}{0.097}[/tex]
Molar mass, MM = 47.42 g/mol.
MM of [tex]NO_2[/tex] = 14 + 32 = 46 g/mol.
Therefore, a gas with a molar mass of 47.42 g/mol is likely to be [tex]NO_2[/tex] because the molar mass of [tex]NO_2[/tex] is 46 g/mol.
