Correct Answer: Option C

Explanation

Given:

m = 22g

M.V = 22.4dm\(^3\)

V = ?

First, calculate the number of moles n, by dividing mass of CO\(_2\) by the molar mass of CO\(_2\), since  n = \(\frac{mass}{Molar mass}\)

Molar mass of CO\(_2\) = 12 + (2 x 16) = 44g/mol

So, n = \(\frac{mass}{Molar mass}\)

      n = \(\frac{22}{44}\) = 0.5 mole

Also recall that number of moles of gases,  n = \(\frac{Volume}{Molar Volume}\)

               n = \(\frac{V}{M.V}\)

             0.5 = \(\frac{V}{22.4}\)

             V = 0.5 X 22.4

             V = 11.2 dm\(^3\)

Alternatively, 

Given the mass of CO\(_2\) = 22g

Molar mass of CO\(_2\) = 12 + (16X2) = 44g

number of moles, n = \(\frac{m}{M}\) =  \(\frac{22}{44}\) = 0.5 mole.

If  1 mole of CO\(_2\) = 22.4dm\(^3\)

    0.5 mole of CO\(_2\) = X dm\(^3\)

X = 0.5 x 22.4 = 11.2 dm\(^3\)

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