Question

How much volume would 63.9 g of carbon monoxide take up?

Answer 1

Hint: The concept of molar volume will help us to solve this problem quickly. According to the mole concept described; a mole of any substance takes up the volume of 22.4 L at STP.

Complete answer:
Let us solve the given problem stepwise;
Given that-
Mass of CO = 63.9 g
Molar mass of CO = 28 g/mol
Thus, according to the definition of number of moles i.e. ratio of given mass to the molar mass.
Number of moles of CO $=\dfrac{63.9}{28}=2.28moles$
Now,
At STP, we know that molar volume is 22.4 L/mol. Thus,
Volume that CO will take up = Number of moles $\times $ Molar volume
$\begin{align}
  & V=2.28moles\times 22.4L/moles \\
 & V=51.072L \\
\end{align}$

Note:
Do note that one should be firm to understand the logic behind the given question as it is purely based on general mole concepts. One can make a mistake while solving this if solved by the method of molar mass and density; as none of the articles can mention perfect density for gases.