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Question: Calculate the density of \( S{O_2} \) , \( {H_2}S \) , \( N{H_3} \) (in g/litre) at STP....

Calculate the density of SO2S{O_2} , H2S{H_2}S , NH3N{H_3} (in g/litre) at STP.

Explanation

Solution

Hint : Density can be stated as mass per unit volume of a particular substance. It is well known that all gases occupy the same volume on a per mole basis so the density of any specific gas depends upon its molar mass. Therefore, it can be said that the gas possessing a small molar mass tends to have a low density in comparison to a gas possessing a large molar mass.

Complete Step By Step Answer:
STP stands for Standard Temperature and Pressure. At standard conditions, standard temperature (T) equals 0 °C (i.e. 273.15 K), standard pressure (P) equals 1 Atm or 101.3kPa or 760 mmHg. We can say that STP refers to the "standard" conditions that are most often used in the measurement of gas density and volume. At STP, we can use the ideal gas equation to calculate the density of gas in g/litre:
PV=nRTPV = nRT
At STP:
\begin{array}{*{20}{l}} {P = 1{\text{ }}atm} \\\ {T = 273{\text{ }}K} \\\ {R = gas{\text{ }}constant = 0.0821{\text{ }}L{\text{ }}atm{\text{ }}mo{l^{ - 1}}{K^{ - 1}}} \end{array}
Number of moles, ‘n’ can be calculated using the following formula:
n=mMn = \dfrac{m}{M}
m = given mass
M = molar mass
Density of a gas can be calculated as the ratio of given mass and volume.
Substituting this in the ideal gas equation, we get:
PV=mMRT PM=mVRT d=mV ∴PM=dRT \begin{gathered} PV = \dfrac{m}{M}RT \\\ PM = \dfrac{m}{V}RT \\\ d = \dfrac{m}{V} \\\ \therefore PM = dRT \\\ \end{gathered}
Using this formula, we will calculate the density of SO2S{O_2} , H2S{H_2}S , NH3N{H_3} (in g/litre) at STP one by one as follows:
(i) SO2S{O_2}
Molar mass (M) = 32+(2×16)=6432 + (2 \times 16) = 64
Substituting the values, we get:
1atm×64g/mol=d×0.0821Latmmol−1K−1×273K d=2.86g/L \begin{gathered} 1atm \times 64g/mol = d \times 0.0821Latmmo{l^{ - 1}}{K^{ - 1}} \times 273K \\\ d = 2.86g/L \\\ \end{gathered}
(ii) H2S{H_2}S
Molar mass (M) = (2×1)+32=34(2 \times 1) + 32 = 34
Substituting the values, we get:
1atm×34g/mol=d×0.0821Latmmol−1K−1×273K d=1.51g/L \begin{gathered} 1atm \times 34g/mol = d \times 0.0821Latmmo{l^{ - 1}}{K^{ - 1}} \times 273K \\\ d = 1.51g/L \\\ \end{gathered}
(iii) NH3N{H_3}
Molar mass (M) = 14+(3×1)=1714 + (3 \times 1) = 17
Substituting the values, we get:
1atm×17g/mol=d×0.0821Latmmol−1K−1×273K d=0.758g/L \begin{gathered} 1atm \times 17g/mol = d \times 0.0821Latmmo{l^{ - 1}}{K^{ - 1}} \times 273K \\\ d = 0.758g/L \\\ \end{gathered}

Note:
Alternatively, density of a gas can be calculated as the ratio of molar mass and molar volume at STP. The following formula can be used to calculate the density of gas:
D=MVD = \dfrac{M}{V}
D = density of gas at STP
M = molar mass
V = molar volume of a gas at STP (22.4 L/mol)