The volume of $2.8 g$ of carbon monoxide at $27^{o} C$ and $0.821 a t m$ pressure is:

0.03 L

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

3 L

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

0.3 L

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

1.5 L

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

Open in App

Solution

The correct option is D $3 L$
The volume of carbon monoxide= $\frac{n R T}{P}$
Mass of carbon\quad monoxide $= 2.8 g$
$P = 0.821 a t m,$
$T = 27 + 273 = 300 K$
$R = 0.0821$
because gram molecular weight of $C O = 12 + 16 = 28$
because number of moles in 2.8g of $C O = \frac{2.8}{28} = 0.1$
therefore volume $= \frac{0.1 \times 0.0821 \times 300}{0.821} = 3 l i t r e$

Suggest Corrections

Similar questions

27^{o} C

=

m o l^{- 1}

^{- 1}

2.8 g

27^{\circ} C

0.821

27^{o} C

27^{\circ} C

0.821

(R = 0.0821 l i t . a t m m o l^{- 1} K^{- 1})

298 K

- 393.5

- 283.0 k J m o l^{- 1}

Join BYJU'S Learning Program