‘x’ lit of Nitrogen at NTP contains 3.0×1022 molecules. The number of molecules in x / 2 lit of Ozone at NTP are

1 mole of N2=6×1023 molecules of N2=22.4 lit of N23×1022 molecule→? lit of N2x=3×1022×22.46×1023x=1.12 lit of N2No. of molecules in x2 lit of Ozone at N.T.P:=x2lit of O3=1.122lit O3=0.56 lit of O31 mole of O3→22.4 lit O3→6×1023molecule of O30.56 lit O3→x molecule of O3=0.56×6×102322.4=0.15×1023=1.5×1022molecules

‘x’ lit of Nitrogen at NTP contains 3.0×1022 molecules. The number of molecules in x / 2 lit of Ozone at NTP are

1 mole of N2=6×1023 molecules of N2=22.4 lit of N23×1022 molecule→? lit of N2x=3×1022×22.46×1023x=1.12 lit of N2No. of molecules in x2 lit of Ozone at N.T.P:=x2lit of O3=1.122lit O3=0.56 lit of O31 mole of O3→22.4 lit O3→6×1023molecule of O30.56 lit O3→x molecule of O3=0.56×6×102322.4=0.15×1023=1.5×1022molecules

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‘x’ lit of Nitrogen at NTP contains $3.0 \times 10^{22}$ molecules. The number of molecules in x / 2 lit of Ozone at NTP are

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$1.5 \times 10^{22}$

$3.0 \times 10^{22}$

$1.5 \times 10^{20}$

$1.5 \times 10^{21}$

answer is B.

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Detailed Solution

$\begin{array}{l} {1 mole of N}_{2} = 6 \times 10^{23} m o l e c u l e s o f N_{2} = 22.4 {lit of N}_{2} \\ 3 \times 10^{22} molecule \to ? {lit of N}_{2} \\ x = \frac{3 \times 10^{22} \times 22.4}{6 \times 10^{23}} \\ x = 1.12 {lit of N}_{2} \end{array}$

No. of molecules in $\frac{x}{2}$ lit of Ozone at N.T.P:

$\begin{array}{l} = \frac{x}{2} {lit of O}_{3} \\ = \frac{1.12}{2} {lit O}_{3} \\ = 0.56 {lit of O}_{3} \end{array}$

$\begin{array}{l} 1 m o l e o f O_{3} \to 22.4 {lit O}_{3} \to 6 \times 10^{23} {molecule of O}_{3} \\ 0.56 {lit O}_{3} \to x {molecule of O}_{3} \\ = \frac{0.56 \times 6 \times 10^{23}}{22.4} = 0.15 \times 10^{23} = 1.5 \times 10^{22} molecules \end{array}$