Chemistry, 02.12.2019 22:31, aprilreneeclaroxob0c

50 ml of a chlorofluorocarbon gas effused through a porous barrier in 157 seconds. the same volume of argon effused in 76 seconds under the same conditions. what is the formula of the chlorofluorocarbon?

Answers: 2

Show answers

Answers

answered: joshbolaadebawore

$CCl_{2}FCF_{3}$

Explanation:

To solve this equation we can use Graham's Law of effusion to calculate the molar mass of the chloroflourocarbon

$\sqrt{\frac{M_{1} }{M_{2} } } = \frac{t_{1} }{t_{2} } \\\\\sqrt{\frac{M_{1} }{39.948 } } = \frac{157 }{76}\\\\\\M_{1} = 170.477\\$

Which chloroflorocarbon has this molar mass?

C2Cl2F4

answered: Guest

it is stable as it is in group 0 which have full outer shells of electrons

answered: Guest

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.

$m = \frac{moles (solute)}{l ( solution)}$

$1 m = \frac{moles (cacl_{2})}{1 l }= 1 mol cacl_{2}$

let us convert this to grams of cacl₂.

molar mass of cacl2 is 111 g/mol

grams of cacl₂ = $1 mol \times \frac{111g}{mol} = 111 g cacl_{2}$

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 = $\frac{1g}{ml} \times 1000 ml = 1000 g$

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

$molality = \frac{mol (solute)}{kg ( solvent)}$

$molality = \frac{1mol}{0.889kg}$

molality = 1.12 m

step 2 : find boiling point elevation.

the elevation in boiling point is calculated as

$\bigtriangleup t_{b} = i \times k_{b} \times m$

here δtb = boiling point elevation

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

cacl₂ dissociates as $cacl_{2} \rightarrow ca^{2+} + 2 cl^{-}$

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

$\bigtriangleup t_{b} = 3 \times 0.512 \times 1.12$

$\bigtriangleup t_{b} = 1.7$

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

$\bigtriangleup t_{f} = i \times k_{f} \times m$

kf is constant for water and its value is 1.86

$\bigtriangleup t_{f} = 3 \times 1.86 \times 1.12$

$\bigtriangleup t_{f} = 6.2$

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

Other questions on the subject: Chemistry

Chemistry, 22.06.2019 11:30, ayoismeisjjjjuan

Which statement best describes the flow of energy in this scenario

Answers: 1

continue

Chemistry, 22.06.2019 14:00, leahstubbs

8.98 dm3 of hydrogen gas is collected at 38.8 °c. find the volume the gas will occupy at -39.9 °c if the pressure remains constant.

Answers: 3

continue

Chemistry, 23.06.2019 01:30, kcarstensen59070

Which statement justifies that hydrogen peroxide (h2o2) is a polar molecule? the o – h bond is nonpolar and the molecule is asymmetric. the o – h bond is nonpolar and the molecule is symmetric. the o – h bond is polar and the molecule is asymmetric. the o – h bond is polar and the molecule is symmetric.

Answers: 1

continue

Chemistry, 23.06.2019 03:40, ElegantEmerald

The following questions a24 - a26 relate to 100 ml of 0.0150 m solution of benzoic acid (c6h3cooh). ka(c6h3cooh) = 6.4 x 10^-5. what is the ph of the solution after the addition of 1 x 10^-3 moles of naoh? you may assume no volume change to the solution upon addition of the naoh.

Answers: 2

continue

Do you know the correct answer?

50 ml of a chlorofluorocarbon gas effused through a porous barrier in 157 seconds. the same volume o...