Question

A 0.98 gram sample of a volatile liquid was heated to 348 k. the gas occupied 265 ml of space at a pressure of 0.95 atm. what is the molecular weight of this gas?

Answer 1

The molecular weight is
`Z = 111.2 \ g/mol`

Step-by-step explanation:

From the question we are told that

The mass of the sample is
`m = 0.98 \ g`

The temperature is
`T = 348 K`

The volume which the gas occupied is
`V = 265 \ ml = 265 *10^(-3) L`

The pressure is
`P = 0.95 \ atm`

Generally from the ideal gas equation we have that

`PV = n RT`

Here n is the number of moles of the gas while the R is the gas constant with value
`R = 0.0821 \ atm \cdot L \cdot mol^(-1) \cdot K^(-1)`

`n = (PV)/( RT)`

=>
`n = ( 0.95 * 265 *10^(-3) )/( 0.0821 * 348)`

=>
`n = 0.00881 \ mol`

Generally the molecular weight is mathematically represented as

`Z = (m)/(n)`

=>
`Z = (0.98 )/(0.00881)`

=>
`Z = 111.2 \ g/mol`