Calculation based on molar volume — lesson. Science State Board, Class 10.

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1.	Calculate the volume occupied by \(2.5\) \(moles\) of \(CO_2\) at STP.

Data:

Given number of moles of \(CO_2=2.5\)

Molar volume of \(CO_2\) at STP \(=22.4\) \(L\)

Solution:

\begin{array}{l} Number of moles of {CO}_{2} & = & \frac{Volume of {CO}_{2} at STP}{Molar volume at STP} \\ 2.5 & = & \frac{Volume of {CO}_{2} at STP}{22.4} \\ Volume of {CO}_{2} at STP & = & 22.4 \times 2.5 \\ = & 56 L \end{array}

Therefore, the volume occupied by \(2.5\) \(mole\) of \(CO_2\) at STP \(=56\) \(L\)

2.	Calculate the volume occupied by \(12.046\times10^{23}\) of ammonia gas molecules.

Data:

Given the number of ammonia gas molecules \(=12.046\times10^{23}\)

Avogadro’s number \(=6.023\times10^{23}\)

Solution:

\begin{array}{l} Number of moles & = & \frac{Number of molecules}{Avogadro " s number} \\ = & \frac{12.046 \times 10^{23}}{6.023 \times 10^{23}} \\ = & 2 moles \end{array}

\begin{array}{l} Volume occupied by {NH}_{3} & = & Number of moles \times Molar volume \\ = & 2 \times 22.4 \\ = & 44.8 L \end{array}

Therefore, the volume occupied by \(12.046\times10^{23}\) of ammonia gas molecules \(=44.8\) \(L\)

3.	Calculate the volume occupied by \(14\) \(g\) of nitrogen gas.

Data:

Given mass of nitrogen \(=14\) \(g\)

Atomic mass of nitrogen \(=28\)

Solution:

\begin{array}{l} Number of moles & = & \frac{Given mass of the element}{Atomic mass of the element} \\ = & \frac{14}{28} \\ = & 0.5 mole \end{array}

\begin{array}{l} Volume occupied by N_{2} at STP & = & No . of moles \times Molar volume \\ Mass & = & 0.5 \times 22.4 \\ = & 11.2 L \end{array}

Therefore, the volume occupied by \(14\) \(g\) of nitrogen gas \(=11.2\) \(L\)

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7. Calculation based on molar volume