The volume of 2.8 g of carbon monoxide at 27°C and 0.821 atm pressure is(R = 0.0821 lit. atm. mol-1 K-1)

Question

The volume of 2.8 g of carbon monoxide at 27°C and 0.821 atm pressure is(R = 0.0821 lit. atm. mol-1 K-1) (A) 30 L (B) 3 L (C) 0.3 L (D) 1.5 L

Tardigrade Admin · Accepted Answer

Given m = 2.8 g, T = 27°, C = 300K and P = 0.821 atm n = (m/M) = (2.8/28) = 0.1 According to gas equation PV = nRT or V = (0.0821 × 300/0.1 × 0.821) = 3 L

Q. The volume of 2.8 g of carbon monoxide at 27 $^{\circ}$ C and 0.821 atm pressure is $(R = 0.0821 l i t . a t m . m o l^{- 1} K^{- 1})$

30 L

3 L

0.3 L

1.5 L

Given m = 2.8 g, T = 27 $^{\circ}$ , C = 300K and P = 0.821 atm
n = $\frac{m}{M} = \frac{2.8}{28} = 0.1$
According to gas equation
PV = nRT or
V = $\frac{0.0821 \times 300}{0.1 \times 0.821}$ = 3 L

Q. The volume of 2.8 g of carbon monoxide at 27∘C and 0.821 atm pressure is(R=0.0821lit.atm.mol−1K−1)

30 L

3 L

0.3 L

1.5 L

Given m = 2.8 g, T = 27∘, C = 300K and P = 0.821 atm n = Mm​=282.8​=0.1 According to gas equation PV = nRT or V = 0.1×0.8210.0821×300​ = 3 L

Q. The volume of 2.8 g of carbon monoxide at 27 $^{\circ}$ C and 0.821 atm pressure is $(R = 0.0821 l i t . a t m . m o l^{- 1} K^{- 1})$

Given m = 2.8 g, T = 27 $^{\circ}$ , C = 300K and P = 0.821 atm
n = $\frac{m}{M} = \frac{2.8}{28} = 0.1$
According to gas equation
PV = nRT or
V = $\frac{0.0821 \times 300}{0.1 \times 0.821}$ = 3 L