What is the molar mass of a gas if 1.80 g occupy 590. mL at 735 mm Hg and 74.0°C?

To do this we need to use the equation PV=nRT P=Pressure (atm) V=Volume (Liters) n= Number of moles (mols) R=Constant (.0821 L * atm/ mol * K) T=Temperature (Kelvin) First we need to get all the number into the correct units 735 mmHg * (1 atm / 760 mmHg) = .967 atm 590 mL ---> .590 L 74.0°C +273 = 347 K .967 atm * .590 L = n * (.0821 L * atm/ mol * K) * 347 K Solving for n we get: n = 0.0200 mols Now we can set up a proportion ( 1.80 g / .0200 mols ) = ( x g / 1 mol ) Solving for x gives us 90 g / mol

P=dRT/mm. mm=dRT/P P=735 mmHg R=.08206 atm*L/(mol*K) or 8.314 kPam*L/(mol*K) T=347K D=M/V=1.8/.590=3.051 Because pressure is in mmHg you must convert it to atm or kPa. I'm going to use atm P=735mmHg=(735/760)atm= .9671 mm=(3.051)(.08206)(347)/(.9671)=89.832…

To do this we need to use the equation PV=nRT P=Pressure (atm) V=Volume (Liters) n= Number of moles (mols) R=Constant (.0821 L * atm/ mol * K) T=Temperature (Kelvin) First we need to get all the number into the correct units 735 mmHg * (1 atm / 760 mmHg) = .967 atm 590 mL ---> .590 L 74.0°C +273 = 347 K .967 atm * .590 L = n * (.0821 L * atm/ mol * K) * 347 K Solving for n we get: n = 0.0200 mols Now we can set up a proportion ( 1.80 g / .0200 mols ) = ( x g / 1 mol ) Solving for x gives us 90 g / mol

Ideal gas law question, atleast i hope it is. This is from an earlier chemistry class. PV=nRT where P is the pressure in kPA V is the volume in L n is moles R is rate (8.314) T temperature in kelvin Isolate n, then divide m/n to get M M= 89.9 g/mol (sig digs)

P=dRT/mm. mm=dRT/P P=735 mmHg R=.08206 atm*L/(mol*K) or 8.314 kPam*L/(mol*K) T=347K D=M/V=1.8/.590=3.051 Because pressure is in mmHg you must convert it to atm or kPa. I'm going to use atm P=735mmHg=(735/760)atm= .9671 mm=(3.051)(.08206)(347)/(.9671)=89.832...