Question

what will be the volume occupied by 100 grams of oxygen gas at stp?
a 35 l
b 70 l
c 100 l
d 22 l

Explanation:

Step1: Find molar mass of \(O_2\)

Molar mass of \(O\) is \(16\space g/mol\), so for \(O_2\), it's \(2\times16 = 32\space g/mol\).

Step2: Calculate moles of \(O_2\)

Moles \(n=\frac{\text{mass}}{\text{molar mass}}=\frac{100\space g}{32\space g/mol}\approx3.125\space mol\).

Step3: Use molar volume at STP

At STP, 1 mole of gas occupies \(22.4\space L\). So volume \(V = n\times22.4\space L/mol\).
\(V = 3.125\space mol\times22.4\space L/mol = 70\space L\).

Answer:

b. 70 L