find the volume, in liters, of 980 mg of CS2(s) at STP?
i need EXPLANATIONS of how to do this stuff! <3
- Dr WLv 71 decade agoBest Answer
are you sure about this question?
CS2 is carbon disulfide. it has a melting point of -112C and a boiling point of +46C at P = 1 atm. T in STP is 0 C... so CS2 is a liquid at STP.. you wrote CS2(s)...and the other answerer used the ideal gas law.. CS2(g)...
CS2(l) has a density of 1.266 g/mL at 25C.. not sure what it is at 0 C (STP).. but it's probably close to 1.266 g/mL...
980 mg x (1 g / 1000 mg) x (1 mL / 1.266g) = 1.241 mL...again at 25C
- steve_geo1Lv 71 decade ago
Atomic weights: C=12 S=32 CS2=76
980 mg = 0.980g
0.980gCS2 x 1molCS2/76gCS2 x 22,400mLCS2/1molCS2 = 289 mL CS2 to three significant figures
22,400 mL = 22.4 L is the molar volume of any perfect gas at STP.
- Doc89891Lv 71 decade ago
Use the ideal gas equation: PV = nRT
P is the pressure in atmospheres
V is the volume in liters
n is the number of moles of gas
R is the gas constant: 0.08205 liter-atm/mole degree Kelvin
T is the temperature in Kelvin
STP is 0 degrees Celsius, 1 atmosphere.
Convert to Kelvin: degrees K = 0 C +273 = 273 K
V = nRT/P = (n X0.08205 X 273)/ 1atm
The number of moles, n, will be the grams of CS2 divided by the mole weight of CS2 - that is
0.980 grams/(12+32+32) = 0.98g/76 g/mole = 0.01289 moles of CS2