PV=nRTR = 0.08205746 Latm/kmol
20C + 273.15 = 293.15K
n = (0.090 ATM * 1.8 L) / ( 0.08205746 Latm/kmol * 293.15K)
n = 0.00673mol
589g / 0.00673 mol
87459.75 g/mol
well this is how you do it
now this is an unbelievable high molar mass so check the values of mass and pressure and you may need to recompute
When the unknown liquid is heated and turned into vapor, the unknown will not occupy the whole container. In the equation to find the molar mass (nRT)/PV, the volume will be greater than the actually volume of the unknown, thus leaving a molar mass that is less than the actual molar mass
0.52 mol.
C6h12o3
Molar gas volume is the volume of ONE moel of gas. It only depends on the pressure and temperature, not on the kind of gas. Molar volume at standard temperature and standard pressure is always 22,4 Litres (for any gas)
Find the constant, which i think is 3.7*10^-4 and multiply it by 1.0. and you should get your answer
When the unknown liquid is heated and turned into vapor, the unknown will not occupy the whole container. In the equation to find the molar mass (nRT)/PV, the volume will be greater than the actually volume of the unknown, thus leaving a molar mass that is less than the actual molar mass
The molar mass is 23,33 g.
rams of X2O3 gas makes 2 atm pressure at 273 C in a 8.96 liter container. So, howmany g.mol is the molar mass of the X atom of the X atom (O:16)?
molar mass of unknown/molar mass of empirial = # of empirical units in the molecular formula. Example: empirical formula is CH2O with a molar mass of 30. If the molar mass of the unknown is 180, then 180/30 = 6 and molecular formula will be C6H12O6
0.52 mol.
C6h12o3
C6h12o3
Molar gas volume is the volume of ONE moel of gas. It only depends on the pressure and temperature, not on the kind of gas. Molar volume at standard temperature and standard pressure is always 22,4 Litres (for any gas)
MolarMass = [density x gas constant x temperature(in kelvin)] / pressure (in atm)
The molecules of the solvent are combining, cutting the concentration of the solvent in half. This, therefore, doubles the molar mass of the solute. Example: 1. Suppose the freezing point of a solution of 2.00 g of an unknown molecular substance in 10.00 g of the solvent benzene is measured. If the solution freezes at a temperature of 6.33°C lower than pure benzene itself, calculate the molar mass of the unknown substance. - 6.33/5.12= 1.24m, solve for x- 1.24m= x/0.01 kg, x=0.0124 mol, 2.00 g/0.0124mol=161.29 g- your molar mass - now cut the amount of solvent, 0.01Kg(10.00g), in half 1.24m=x/0.005 kg, x=0.0062 mol, 2.00 g/0.0062mol = 322.581 g, your larger molar mass
I presume that molar mass is needed n = pV/RT = 1 atm x 0.250 L/ 0.08206 x 273 K= 0.0112 molar mass = 2.50/ 0.0112 =224 g/mol total moles = 5.0 3.5 / 5.0 = partial pressure CO2 / 7.0 partial pressure CO2 = 4.9 atm Goodluck from michigan tech - Charlesworth sucks
Molar concentration is defined as the amount of a gas divided by the volume of gas. According to the gas identity, at standard temperature and pressure, even if the amount of the gas is constant, the volume of gas changes. Thus, the molar concentration changes depending on the gas identity.