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JEE Main levelSingle correct optionDifficulty 2 of 3ig068

The number of molecules in $1.12 mL$ of an ideal gas at STP (molar volume $22.4 L$ , $N_{A} = 6.02 \times 1 0^{23}$ ) is

A $3.01 \times 1 0^{19}$
B $3.01 \times 1 0^{20}$
C $6.02 \times 1 0^{19}$
D $3.01 \times 1 0^{22}$

Answer: A

Moles

= \frac{1.12 \times 1 0 ^{- 3}}{22.4} = 5.0 \times 1 0^{- 5} mol

. Molecules

= 5.0 \times 1 0^{- 5} \times 6.02 \times 1 0^{23} = 3.01 \times 1 0^{19}

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