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A gas filled in a bulb of capacity 25.1 ...

A gas filled in a bulb of capacity 25.1 mL at 27°C and 750 mm pressure weighs 0.072 g. If 1 litre of hydrogen at S.T.R weighs 0.09 g, calculate the molecular mass of the gas.

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Let the volume of the gas at STP be `V_(2)` mL
`P_(1) = 750 mm, V_(1) = 25.1 mL, T_(1) = 27^(@) C = 300 K`
`P_(2) = 760 mm, V_(2) = ?, T_(2) = 273 K`
According to the gas equation:
`(P_(1)V_(1))/T_(1) = (P_(2)V_(2))/T_(2)` or `V_(2) = (P_(1)V_(1)T_(2))/(T_(1)P_(2))`
Substituting the values, we get
`V_(2) = (750 xx 25.1 xx 273)/(300 xx 760)`
= 22.5 mL
Mass of 22.5 mL of `H_(2)` at S.T.P.
`=0.09/1000 xx 22.5 = 2.025 xx 10^(-3) g`
V.D. = `("Mass of 22.5 mL of gas at S.T.P")/("Mass of 22.5 mL of H_(2) at S.T.P")`
`=0.072/(2.025 xx 10^(-3))`
= 35.55
Hence, molecular mass `=2 xx V.D. = 2 xx 35.55 = 71.1`
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