The number of chlorine atoms in 20 mL of chlorine gas at STP is x- 1021- The value of x Rounded off to the Nearest integer is -Assume chlorine is an ideal gas at STP- NA-6-023- 1023-

Moles of Cl_2 in 20 mL of it at STP =2022400=8.9× 10^ 4 mol Thus, no. of Cl atoms =moles× N_A× atomicity ..... eq. (i) Atomicity is the number of atoms per mole ...

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Standard XII

Chemistry

Avogadro's Law & Gay Lussac's Law

The number of...

Question

The number of chlorine atoms in 20 mL of chlorine gas at STP is $x \times 10^{21}$ . The value of x (Rounded off to the Nearest integer) is
.
[Assume chlorine is an ideal gas at STP,
$N_{A} = 6.023 \times 10^{23}$ ]

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Solution

Moles of $C l_{2}$ in $20 m L$ of it at STP $= \frac{20}{22400} = 8.9 \times 10^{- 4} m o l$
Thus, no. of $C l$ atoms $= m o l e s \times N_{A} \times a t o m i c i t y$ ..... eq. (i)
Atomicity is the number of atoms per molecule of a species.
$∴$ Atomicity of $C l_{2} = 2$
Putting values in eq(i), we get;
no. of $C l$ atoms $= 8.9 \times 10^{- 4} \times 6.022 \times 10^{23} \times 2 \approx 1 \times 10^{21}$
Thus value of $x = 1$ .

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The number of chlorine atoms in 20 mL of chlorine gas at STP is x×1021. The value of x (Rounded off to the Nearest integer) is . [Assume chlorine is an ideal gas at STP, NA=6.023×1023]

The number of chlorine atoms in 20 mL of chlorine gas at STP is $x \times 10^{21}$ . The value of x (Rounded off to the Nearest integer) is
.
[Assume chlorine is an ideal gas at STP,
$N_{A} = 6.023 \times 10^{23}$ ]