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112 cm^(3) at S.T.P. of a gaseous fluori...

112 `cm^(3)` at S.T.P. of a gaseous fluoride of phosphorus has a mass of 0.63 g calculate the relative molecular mass of fluoride. If the molecule of the fluoride contains only one atom of phosphorus, determine the formula of the phosphorus fluoride. [F = 19, P = 31]

Text Solution

Verified by Experts

Molar mass of the gase `(22400 cm^(3))/(112 cm^(3))xx0.63g=126g`
Determination of formula: Let us suppose that the formula of the fluoride is `PF_(x)`
Formula mass of `PF_(x) = P + XF = 31 + x xx 19`
Molecular mass of `PF_(x) = 126`
Now, Formula mass = Molecular mass
Or, `31+19x=16`
And `x=(126-31)/(29)=5`
Formula of phosphorus fluoride `PF_(x)=PF_(5)`
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