Calculate the volume in litres of 49.8 gram of HCl gas at STP. ( molar mass of HCl= $36.5 g - m o l^{- 1}$ )

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Solution

Molar mass of $H C l = 36.5 g = 1$ mole
$∴ 36.5 g$ occupies $22.4 d m^{3}$ of volume
$∴$ $49.8 g$ of $H C l$ occupies $\frac{49.8 \times 22.4}{36.5}$
$= 30.56 d m^{3}$
$= 30.56 L$ .

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7.5 grams of a gas occupy 5.8 litres of volume at STP the gas is

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Calculate the volume in litres of 49.8 gram of HCl gas at STP. ( molar mass of HCl=36.5g−mol−1)

Molar mass of HCl=36.5g=1 mole∴36.5g occupies 22.4dm3 of volume∴ 49.8g of HCl occupies 49.8×22.436.5 =30.56dm3 =30.56L .

Calculate the volume in litres of 49.8 gram of HCl gas at STP. ( molar mass of HCl= $36.5 g - m o l^{- 1}$ )

Molar mass of $H C l = 36.5 g = 1$ mole
$∴ 36.5 g$ occupies $22.4 d m^{3}$ of volume
$∴$ $49.8 g$ of $H C l$ occupies $\frac{49.8 \times 22.4}{36.5}$
$= 30.56 d m^{3}$
$= 30.56 L$ .