# The volume of 2.8g of carbon monoxide at and 0.821atm pressure is: (R= 0.0821 lit.atm.mol-1K1):

A.1.5 lit

B.0.3lit

C.3lit

D.30lit

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**Hint:**Ideal gas law is a combined set of gas laws that is a thermodynamic equation that allows us to relate the temperature, volume, and number of moles present in a sample of gas.

Formula used: The formula for calculating V is:

\[{\text{V = }}\dfrac{{{\text{WRT}}}}{{{\text{PM}}}}\]

**Complete step by step answer:**

As we know that, the formula is \[{\text{V = }}\dfrac{{{\text{WRT}}}}{{{\text{PM}}}}\]

Where,

W=weight

R= constant (given in the question)

P= pressure

T=temperature

M= molar mass of carbon monoxide (i.e. 28)

Hence, by applying the formula and putting the values

We get,

\[

V = \dfrac{{2.8 \times 0.0821 \times 300}}{{0.821 \times 28}} \\

= \dfrac{{68.9}}{{22.9}} \\

= 3lit \\

\]

Thus, we get V= 3 lit

**Hence, option C is correct.**

**Note:**

Another way is,

From gas equation, PV=NRT,

\[

N = \dfrac{m}{M} = \dfrac{{2.8}}{{28}} = 0.1 \\

V = \dfrac{{0.821 \times 300}}{{0.1 \times 0.821}} = 3L \\

\]