How many grams of pentane are contained in a vapor sample that occupies a volume of 0.285 l at 57°c and 1.0 atm? .

0.260 g
0.515 g
0.760 g
1.00 g

The answer to your question is: mass = 0.760 g

Explanation:

Data

V = 0.285 L

T = 57°C = 330°K

P = 1 atm

R = 0.082 atm L/mol°K

mass = ?

Formula

                PV = nRT

                n = PV / RT

Substitution

                n = (1)(0.285) / (0.082)(330)

                n = 0.285 / 27.06

                n = 0.011

mW Pentane = (12 x 5) + (12 x 1) = 72 g

                                 72 g of pentane 1 mol

                                   x                         0.0105 mol

                                  x = (0,0105 x 72) / 1

                                  x = 0.756 g ≈ 0.760 g

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"