In a mixture of gases, the volume content of a gas is 0.06 % at STP. Calculate the number of molecules of the gas in 1 L of the mixture.

Question

In a mixture of gases, the volume content of a gas is 0.06 % at STP. Calculate the number of molecules of the gas in 1 L of the mixture. (A) 1.613 ×1023 (B) 6.023 ×1023 (C) 1.61 ×1027 (D) 1.61 ×1019

Tardigrade Admin · Accepted Answer

Volume of gas in 1 L = (0.06/100) = 6× 10-4L Number of molecules of CO2 = n × NA (6×10-4 /22.4) ×6.023 ×1023 = 1.61 ×1019

Q. In a mixture of gases, the volume content of a gas is $0.06%$ at $STP$ . Calculate the number of molecules of the gas in $1 L$ of the mixture.

$1.613 \times 1 0^{23}$

$6.023 \times 1 0^{23}$

$1.61 \times 1 0^{27}$

$1.61 \times 1 0^{19}$

Volume of gas in $1 L = \frac{0.06}{100} = 6 \times 1 0^{- 4} L$
Number of molecules of $C O_{2} = n \times N_{A}$
$\frac{6 \times 1 0 ^{- 4}}{22.4} \times 6.023 \times 1 0^{23}$
$= 1.61 \times 1 0^{19}$

Q. In a mixture of gases, the volume content of a gas is 0.06% at STP. Calculate the number of molecules of the gas in 1L of the mixture.

1.613×1023

6.023×1023

1.61×1027

1.61×1019