How many number of moles of nitrogen will be present in $2.24 L$ of nitrogen gas at $S T P ?$

9.9

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.099

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

0.001

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1.00

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $0.099$
$P V = n R T$

$P = 1 a t m, T = 273 K, V = 2.24 L$

$R = 0.0821 L a t m K^{- 1} m o l^{- 1}$

$n = \frac{1 \times 2.24}{0.0821 \times 273} = 0.099$ moles.

Suggest Corrections

Similar questions

The number of moles present in 224 L of nitrogen at STP is

N H_{3}

Join BYJU'S Learning Program