Chemistry, 25.02.2020 02:31 mroueh21

A solution is prepared by mixing 2.17 g of an unknown non-electrolyte with 225.0 g of chloroform. The freezing point of the resulting solution is –64.2oC. The freezing point of pure chloroform is – 63.5oC and its kf = 4.68oC m-1. What is the molecular mass of the unknown?

Answers: 2

Show answers

Answers

Answer from: oomale

The molar mass of the unknown non-electrolyte is 64.3 g/mol

Explanation:

Step 1: Data given

Mass of an unknown non-electrolyte = 2.17 grams

Mass of chloroform = 225.0 grams

The freezing point of the resulting solution is –64.2 °C

The freezing point of pure chloroform is – 63.5°C

kf = 4.68°C/m

Step 2: Calculate molality

ΔT = i*kf*m

⇒ ΔT = The freezing point depression = T (pure solvent) − T(solution) = -63.5°C + 64.2 °C = 0.7 °C

⇒i = the van't Hoff factor = non-electrolyte = 1

⇒ kf = the freezing point depression constant = 4.68 °C/m

⇒ m = molality = moles unknown non-electrolyte / mass chloroform

0.7 °C = 1 * 4.68 °C/m * m

m = 0.150 molal

Step 3: Calculate moles unknown non-electrolyte

molality = moles unknown non-electrolyte / mass chloroform

Moles unknown non-electrolyte = 0.150 molal * 0.225 kg

Moles unknown non-electrolyte = 0.03375 moles

Step 4: Calculate molecular mass unknown non-electrolyte

Molar mass = mass / moles

Molar mass = 2.17 grams / 0.03375 moles

Molar mass = 64.3 g/mol

The molar mass of the unknown non-electrolyte is 64.3 g/mol

Answer from: Quest

the cold front replaces the warmer air at the ground level changing the wind to be breezy and denser.there is also an immediate drop in the temperature.

Answer from: Quest

answer : (a) volume of the gas = 8.67 × $10^{-3}$ l

(b) second virial coefficient b = -1.21226 l/mole

solution : given, t = 300k

p = 20atm

z = 0.86

n = 8.2 mmoles = 8.2 × $10^{-3}$ moles

r = 0.082 l atm/k/mol

find, (a) volume of gas and (b) second virial coefficient b at 300k.

formula used : $z=\frac{pv}{nrt}$

and $z=1+\frac{b}{v}$

(a) $z=\frac{pv}{nrt}$

$v=\frac{znrt}{p}$

= $\frac{0.86\times 8.2\times[tex]10^{-3}$ \times 0.082\times 300}{20}[/tex]

= 8.67 × $10^{-3}$ l

(b) $z=1+\frac{b}{v}$

$b=(z-1)v$

= (0.00867-1) × 8.659

= - 8.5839 l/mole

Another question on Chemistry

Chemistry, 21.06.2019 17:00

This line graph compares the growth of plants that were kept in the sun for different amounts of time.a) on day 7, the plants kept in the sun for 3 hours were how tall? b) on day 7, the plants kept in the sun for 6 hours were how tall? c) on day 10, the plants kept in the sun for 9 hours were how tall? d) on day 11, the plant that was grown with 1 hour of sunlight was how tall? e) based on the graph, the plant grows best in what amount of sunlight?

Answers: 1

Answer

Chemistry, 22.06.2019 10:40

What is the ph of a 0.0010 m hno3? 1.0 3.0 4.0 5.0

Answers: 2

Answer

Chemistry, 22.06.2019 11:00

Problem page combustion of hydrocarbons such as pentane ( c5 h12 ) produces carbon dioxide, a "greenhouse gas." greenhouse gases in the earth's atmosphere can trap the sun's heat, raising the average temperature of the earth. for this reason there has been a great deal of international discussion about whether to regulate the production of carbon dioxide.(a) write a balanced chemical equation, including physical state symbols, for the combustion of liquid pentane into gaseous carbon dioxide and gaseous water. (b) suppose 0.350 kg of pentane are burned in air at a pressure of exactly 1 atm and a temperature of 20.0 degree c. calculate the volume of carbon dioxide gas that is produced.be sure your answer has the correct number of significant digits.

Answers: 2

Answer

Chemistry, 22.06.2019 11:00

As air becomes more dense, (select all that apply) o. air weighs less o. gas molecules are closer together o. air is colder o. air weighs more o. gas molecules are further apart o. air is hotter

Answers: 3

Answer

You know the right answer?

A solution is prepared by mixing 2.17 g of an unknown non-electrolyte with 225.0 g of chloroform. Th...

Questions