937 AM incloudwebstudycom ATT LTE 1 If PV nRT and n mMm r

9:37 AM incloud.webstudy.com AT&T; LTE 1 ) If PV= nRT and n = m/Mm , rearrange the ideal gas equation to solve for Mm (Molar mass) 2) A 10.67 L sample of gas at 35C and 2.3 atm weighs 20.567 g. What is its molar mass? 3) What would be the value of the universal gas constant R (0.0821 L-atm/mole-K) in mm3-kPa?

Solution

PV   = nRT

no of moles (n) = weight of substance/Gram molar mass

weight of substance = m

Gram molar mass   = M_m

                     n    = m/M_m

   PV = nRT

PV = mRT/M_m

M_m    = mRT/PV

2. V   = 10.67L

    P = 2.3atm

   T   = 35 + 273   = 308K

   m = 20.567g

M_m    = mRT/PV

         = 20.567*0.0821*308/2.3*10.67

         = 21.2g/mole >>>>>answer

3. R   = 0.0821L-atm/mole-K

1L   = 1000000mm^3

1 atm = 101.325KPa

R   = 0.0821*1000000mm^3 *101.325KPa/ mole-K

         = 8318782.5 mm^3-KPa/mole-K    >>>>>answer