Question

1.8 grams of water occupied volume at STP is​

Answer 1

The volume occupied by 1.8 grams of water at STP is 2.24 L

Explanation:

According to ideal gas equation:

$PV=nRT$

P = pressure of gas = 1 atm  (at STP)

V = Volume of gas = ?

n = number of moles of water = $\frac{\text {given mass}}{\text {Molar mass}}=\frac{1.8g}{18g/mol}=0.1mol$

R = gas constant = $0.0821Latm/Kmol$

T = Temperature = $273K$    (at STP)

$V=\frac{nRT}{P}$

$V=\frac{0.1mol\times 0.0820 L atm/K mol\times 273K}{1atm}=2.24L$

Thus volume occupied by 1.8 grams of water at STP is 2.24 L

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Answer 2

Answer:

2.24 l as it contain 0.1 mole

Explanation: