find the volume, in liters, of 980 mg of CS2(s) at STP?

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Answer 1

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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)...

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from here...

http://www.sigmaaldrich.com/catalog/ProductDetail....

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

Answer 2

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.

Answer 3

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