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

Answer:

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

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  $R = 0.0821 \ atm \cdot L \cdot mol^{-1} \cdot K^{-1}$

        $n = \frac{PV}{ RT}$

=>      $n = \frac{ 0.95 * 265 *10^{-3} }{ 0.0821 * 348}$

=>      $n = 0.00881 \ mol$

Generally the molecular weight is mathematically represented as

          $Z = \frac{m}{n}$

=>      $Z = \frac{0.98 }{0.00881}$

=>      $Z = 111.2 \ g/mol$