A sample of argon has a volume of 0.165L at -34.0°C and a 0.98atm of pressure. What would the volume of this gas at STP be?

0.185 L

Explanation:

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

where,

= initial pressure of gas = 0.98 atm

= final pressure of gas = 1 atm  (at STP)

= initial volume of gas = 0.165 L

= final volume of gas = ?

= initial temperature of gas =

= final temperature of gas =  (at STP)

Now put all the given values in the above equation, we get:

Thus the volume of this gas at STP be 0.185 L

The molecular formula for the organic compound is C₈H₈O₂

Explanation:

70.58 % C , 5.9 % H , and 23.50 % O

This is a percent by mass, so:

in 100 g of compound I have 70.58 g of C, 5.9 g of H and 23.5 g of O.

We need to apply the formula for freezing point depression, the colligative property.

ΔT = Kf . m

Where ΔT means T°F (pure solvent) − T°F (solution)

Kf, the cryoscopic constant and m is molality (mol of solute in 1kg of solvent); Kf for cyclohexane is 20.8 °C/m

6.59°C - 3.29°C = 20.8 °C/m . m

3.3°C = 20.8 °C/m . m

3.3°C /20.8 m/°C = molal

0.158 m/kg

In 1kg of solvent I have 0.158 moles, but I have 718.1 g of cyclohexane

1000 g 0.158 moles of solute

718.1 g 0.114 moles of solute

To find out the molar mass of my compound I have to divide

mass/mol → 15.5 g /0.114 m = 136.04 g/m

Now I have to work with the percent and think this rules of three:

100 g of compound has 70.58 g of C __5.9 g of H __23.5 g of O

136.04 g of compound has ___  96.02 g of C __8.02 g of H __ 32 g of O

Molar mass of C = 12 g/m

Molar mass of H = 1 g/m

Molar mass of O = 16 g/m

g / molar mass = moles

C: 96 g / 12 g/m = 8

H: 8 g / 1 g/m =8

O: 32 g/ 16 g/m = 2

The molecular formula for the organic compound is C₈H₈O₂