Back to Course

SS1: CHEMISTRY - 2ND TERM

0% Complete
0/0 Steps
  1. Kinetic Theory of Matter | Week 1
    3 Topics
    |
    1 Quiz
  2. Kinetic Theory of Matter & Gas Laws I | Week 2
    3 Topics
    |
    1 Quiz
  3. Gas Laws II | Week 3
    3 Topics
    |
    1 Quiz
  4. Gas Laws III | Week 4
    3 Topics
    |
    1 Quiz
  5. Gas Laws IV | Week 5
    3 Topics
    |
    1 Quiz
  6. Mole Concept | Week 6
    5 Topics
    |
    1 Quiz
  7. Acid, Bases and Salts I | Week 7
    7 Topics
    |
    1 Quiz
  8. Acid, Bases and Salts II | Week 8
    6 Topics
    |
    2 Quizzes
  9. Acid, Bases and Salts III | Week 9
    3 Topics
    |
    2 Quizzes
  10. Acid, Bases and Salts IV | Week 10
    4 Topics
    |
    1 Quiz
  11. Acid, Bases and Salts V | Week 11
    5 Topics
    |
    1 Quiz

Quizzes




  • Kofa Study
SS1: CHEMISTRY – 2ND TERM Gas Laws IV | Week 5 Diffusion Rates of Ammonia (NH3) and Concentrated Hydrochloric Acid (HCl)
Lesson 5, Topic 3
In Progress

Diffusion Rates of Ammonia (NH3) and Concentrated Hydrochloric Acid (HCl)

Lesson Progress
0% Complete

Topic Content:

  • Diffusion Rates of Ammonia (NH3) and Concentrated Hydrochloric Acid (HCl)
  • Theory Questions and Answers

Experiment: Diffusion Rates of Ammonia (NH3) and Concentrated Hydrochloric Acid (HCl)

(N = 14, H = 1, Cl = 35.5) NH3 = 17, HCl = 36.5

Apparatus:

Glass tube with open ends, cotton wool.

Procedure:

Soak a small quantity of cotton wool in a small solution of Ammonia. Place the soaked cotton wool at one open end of the dry glass tube.

Soak another cotton wool in a small quantity of concentrated hydrochloric acid.

Place it at the other open end of the glass tube.

Screenshot 2024 01 31 at 08.10.06

Observation

 

You are viewing an excerpt of this Topic. Subscribe Now to get Full Access to ALL this Subject's Topics and Quizzes for this Term!

Click on the button "Subscribe Now" below for Full Access!

Subscribe Now

Note: If you have Already Subscribed and you are seeing this message, it means you are logged out. Please Log In using the Login Button Below to Carry on Studying!

Responses

Your email address will not be published. Required fields are marked *

Question 1

(a) state Graham’s Law of Diffusion

Answer –

Graham’s law of diffusion states that at constant temperature and pressure, the rate of diffusion of a gas is inversely proportional to the square root of its density or relative molecular mass.

(b) 60cm3 of Hydrogen gas different through a Porous membrane in 10minutes. The same volume of gas G diffused through the same membrane in 37.4minutes. Determine the Relative molecular mass of G [H = 1 ]

Answer –

:- \( \frac{R_1}{R_2} = \frac{\sqrt{m_2}}{\sqrt{m_1}} \)

⇒\( \frac{RH_1}{R_G} = \frac{\sqrt{m_G}}{\sqrt{mH_2}} \)

At equal volume,

:- \( \frac{t_2}{t_1} = \frac{\sqrt{m_2}}{\sqrt{m_1}} \)

⇒ \( \frac{t_G}{tH_2} = \frac{\sqrt{m_G}}{\sqrt{H_2}} \)

:- \( \frac{37.4}{10} = \frac{\sqrt{m_G}}{\sqrt{2}} \)

Square both sides,

:- \( \frac{(37.4)^2}{(10)^2} = \frac{m_G}{2} \)

:- \( \frac{1,398.78}{100} = \frac{m_G}{2}\)

Cross multiply,

2,797.52 = 100MG

MG  = \( \frac {2,797.52}{100} \)

MG = 27.98

Question 2

(a) Give three conditions under which Real gas behave Ideally 

Answer – 

  1. At low pressure
  2. Under laboratory condition
  3. At high temperature

(b) How many moles of Oxygen gas are there in 350cm3 of the gas at 00C and 380mmHg pressure. (O2 =32, R =0.082 atm dm-3k-1mol-1)

Answer – 

V = 350cm3 ⇒ 0.35dm3

P = 380mmHg

T = 273k

n = ?

R = 0.082 atm dm-3k-1mol-1

n = \( \frac{PV}{RT} \)

n = \( \frac{380 \; \times \; 0.35}{0.082 \; \times \; 273} \)

n = \( \frac{133}{22.39} \)

n = 5.95moles

Question 3

 (a) Show that for an Ideal gas PV=nRT, and define all the terms in the expression.

Answer – 

:- \( \scriptsize P \propto T \)

:- \( \scriptsize P \propto \normalsize \frac{1}{V} \) ………….(2)

:- \( \scriptsize P \propto n \) ………….(2)

:- \( \scriptsize P \propto n  \normalsize \frac{T}{V}\)

:- \( \scriptsize PV \propto nT \) 

:- \( \scriptsize PV \propto KnT \)  K = constant of proportionality

PV = nRT ; where K = R

∴ PV = nRT

 

 (b) Calculate the pressure at 270C OF 16.0g of Oxygen gas occupying 2.50dm3

Answer – 

PV = nRT

P = ?

V = 2.50dm3

n = \( \frac{m}{M} =  \frac{16}{32}  = \scriptsize 0.5\: moles\)

m = 16g

M = 32

R = 8.31Jk-1mol-1

T = 300k

P = \( \frac{nRT}{V} \)

P = \( \frac{0.5 \; \times \; 8.31 \; \times 300}{3.50} \)

P = \( \frac{1246.5}{2.50} \)

= 498.6 PA

Question 5

Arrange the following gases in ascending order of their rates of diffusion   

[S=32, Cl=35.5, H=1, O=16, C=12]                  

(a) Chlorine, Hydrogen, Oxygen, Carbon(IV) Oxide  

Answer – 

:- \(\scriptsize R \propto \normalsize \frac{1}{\sqrt{m}} \)

(Cl) Chlorine = \(\scriptsize R \propto \normalsize \frac{1}{\sqrt{70}} \\ = \scriptsize 0.12 \)

(H2) Hydrogen = \(\scriptsize R \propto \normalsize \frac{1}{\sqrt{2}} \\ = \scriptsize 0.71 \)

(O2) Hydrogen = \(\scriptsize R \propto \normalsize \frac{1}{\sqrt{32}} \\ = \scriptsize 0.18 \)

(CO2) Carbon (IV) Oxide = \(\scriptsize R \propto \normalsize \frac{1}{\sqrt{44}} \\ = \scriptsize 0.15 \)

 

(b) Sulphur(IV) Oxide, Ammonia, Nitrogen  

Answer – 

SO2 = 64;

NH3 = 17;

N2 = 28

SO2 = \( \frac{1}{\sqrt{64}} = \scriptsize 0.13 \)

NH3  = \( \frac{1}{\sqrt{17}} = \scriptsize 0.24\)

N2 = \( \frac{1}{\sqrt{28}} = \scriptsize 0.19\)

 

(c) Methane, Hydrogen Sulphide, Carbon(IV) Oxide, Hydrogen  

Answer – 

CH3 = 15

H2S = 34

CO2 = 44

H2 = 2

CH3 = \( \frac{1}{\sqrt{15}} = \scriptsize 0.25 \)

H2S = \( \frac{1}{\sqrt{34}} = \scriptsize 0.17 \)

CO2 = \( \frac{1}{\sqrt{44}} = \scriptsize 0.15 \)

H2 = \( \frac{1}{\sqrt{2}} = \scriptsize 0.71 \)

CO2 < H2S < CH3 < H2

 

(d) Ethene, ethane ethyne, Water vapour

Answer – 

C2H4 =28;

C2H6 = 30;

C2H2 = 26;

H2O = 18

C2H4 = \( \frac{1}{\sqrt{28}} = \scriptsize 0.19 \)

C2H6 = \( \frac{1}{\sqrt{30}} = \scriptsize 0.18 \)

C2H2 = \( \frac{1}{\sqrt{26}} = \scriptsize 0.20 \)

H2O = \( \frac{1}{\sqrt{18}} = \scriptsize 0.24 \)

Report

Harassment or bullying behavior
Contains mature or sensitive content
Contains misleading or false information
Contains abusive or derogatory content
Contains spam, fake content or potential malware

Block Member?

Please confirm you want to block this member.

You will no longer be able to:

  • See blocked member's posts
  • Mention this member in posts
  • Invite this member to groups
  • Message this member
  • Add this member as a connection

Please note: This action will also remove this member from your connections and send a report to the site admin. Please allow a few minutes for this process to complete.

Report

You have already reported this .
error: Alert: Content selection is disabled!!