A gas occupies 0.418 L at 740 mm of Hg at 27 C. Calculate (i) molecular weight, if gas weighs 3.0 g (ii) New pressure of gas, if the weight of gas is increased to 7.5 g and the temperature become 280 K.
 

Dear student
Please find below the solution to the asked query:
Given P1 = 740mmHg
V1 = 0.418 L
T1 = 300K

​P2 = 760 mmHg
T2 = 273K
​We have to find V2
$$\begin{gathered} \frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2} \\ substituting the values we get: \\ \frac{740\times 0.418}{300}=\frac{760\times V_2}{273} \\ V2 = 0.370L \end{gathered}$$


Volume at STP = 0.370L
From PV= nRT {we solve for obtaining  the value of n}
     n =PV/RT    [At STP. P= 1atm, T=273K, R= 0.082Latm/K mol
        =$$\begin{gathered} \frac{1\times 0.37}{0.082\times 273} \\ =0.017 moles \end{gathered}$$


i) When gas weighs = 3g, the molecular weight is  
$$\begin{gathered} No.of moles = \frac{weight of the gas}{molar mass} \\ 0.017 =\frac{3}{molar mass} \\ Molar mass =176.5g/mol \end{gathered}$$

ii) When weight of the gas increases to 7.5g and yemperature is 280K, presssure will be:
               PV = nRT
​           $PV=\frac{w}{M}RT$   [as no.of moles = weight of gas/molar mass of the gas]
Substituting the values we get:
$$\begin{gathered} P\times 0.37 = \frac{7.5}{176.5}\times 0.082\times 280 \\ P = 2.64 atm \end{gathered}$$



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