A gas sample containing a constant number of gas molecules has a volume of 2.70 L at a constant pressure and a temperature of 25.0o C. What would be the volume (in Liters) of this gas sample at 75.0o C? Round your answer to 3 sig fig

Explanation:

This question asks us find the volume of a gas sample given a change in temperature. Since the pressure remains constant, we only are concerned with the variables of temperature and volume.

We will use Charles's Law. This states the volume of a gas is directly proportional to the temperature of a gas. The formula is:

The gas starts at a volume of 2.70 liters and a temperature of 25.0 degrees Celsius.

The temperature is increased to 75.0 degrees Celsius, but the volume is unknown.

We are solving for the volume at 75 degrees Celsius, so we must isolate the variable V₂.

It is being divided by 75.0 °C. The inverse operation of division is multiplication, so we multiply both sides of the equation by 75.0 °C.

The units of degrees Celsius (° C) cancel.

The original measurements have 3 significant figures, so our answer must have the same. Currently, the answer has 2. If we add another 0, the value of the answer does not change, but the number of sig figs does.

The volume of this gas sample at 75.0 degrees Celsius is 8.10 Liters.

for points guys

the missing options are :

(1) lower boiling point and a lower freezing point(2) lower boiling point and a higher freezing point(3) higher boiling point and a lower freezing point(4) higher boiling point and a higher freezing point

answer : the correct answer is option (3) higher boiling point and a lower freezing point

explanation :

when a non volatile solute is added to a solvent like water, it results in elevation of boiling point.

let us find the increase in boiling point of water when cacl₂ is added to water to make a solution of 1 m

step 1 : find molality of the solution

let us assume we have 1 l of solution.

let us convert this to grams of cacl₂.

molar mass of cacl2 is 111 g/mol

grams of cacl₂ =

mass of solute = 111 g

assuming density of solution same as water which is 1g/ml we have ,

mass of solution = density * volume

mass of solution =

mass of solution = mass of solvent + mass of solute

mass of solute = 1000g - 111g

mass of solute 889 g = 0.889 kg

the molality of the solution can be calculated as

molality = 1.12 m

step 2 : find boiling point elevation.

the elevation in boiling point is calculated as

here δtb = boiling point elevation

i = vant hoff factor which is the number of ions formed by the solute .

cacl₂ dissociates as

it forms 3 ions . therefore,

i = 3

m = molality which is 1.12 m

kb is constant for a given solvent. for water it is 0.512

the boiling point of water increases by 1.7 °c

step 3: find freezing point depression.

when a non volatile solute is added to water, it results in decrease in freezing point.

the formula to find freezing point depression is

kf is constant for water and its value is 1.86

the freezing point of water decreases by 6.2 °c

therefore we can say that, the solution will have a higher boiling point and a lower freezing point