A) An unknown gas effuses 1.73 times faster than krypton. what is the molar mass

Question

A) An unknown gas effuses 1.73 times faster than krypton. what is the molar mass of the gas?

B) Show by Mathimatical equations how one can determine the molar mass of a volatile liquid by measurement of the pressure, volume, temperature, and mass of the liquid?

Trial              Initial [A]                            Initial [B]                     Initial Rate

1                         0.40                                   0.40                            0.480mol/dm3/s

2                   0.40                                        0.80                           1.92 mol/dm3/s

3                     0.20                                   0.40                            0.24 mol/dm3/s

a) write the rate expression for this reaction.

b) calculate the rate constant, k

c) find the rate if [A] becomes 0.50M and [B] becomes 0.25 M (temp is unchanged)

Explanation / Answer

A ) via grahams law,

rate 1 / rate 2 = ?( molar mass 2 / molar mass 1)

let "1" = known gas...Kr.. molar mass = 83.8 g/mole...

so...
(rate 1 / 1.73 rate 1) = ?( molar mass 2 / (83.8 g/mole))

(1 / 1.73) = ?( molar mass 2 / (83.8 g/mole))
(1 / 1.73)

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