
At NTP, density of a gas is $ 0.00045\,g/mL $ . The vapour density and molecular weight will be:
Answer
544.5k+ views
Hint: To solve this, we will use the ideal gas equation at NTP conditions, and then we will calculate the molecular weight from the formula. Later, that value will be used to calculate the value of the vapour density of the gas.
Formula Used:
Ideal gas law is represented by;
$ PV = nRT $
Where, $ P = $ Pressure of the gas, $ T = $ Temperature of the gas
$ V = $ The volume of the gas, $ n = $ No. of moles of gas and $ R = 0.082 $ (Gas Constant)
$ Vapour\,Density = \dfrac{{Weight\,of\,given\,gas}}{{Weight\,of\,{H_2}}} $
$ density(d) = \dfrac{{mass(M)}}{{volume(V)}} $ or $ volume(V) = \dfrac{{mass(M)}}{{density(d)}}.....(a) $ .
Complete Step-by-step Solution
Given, Density of gas $ = 0.00045\,g/mL = 0.45g/L $
AT NTP condition, $ P = 1\,atm $ , $ V = 1\,L $ , $ R = 0.082 $ and $ T = 273\,K $
Now we will use the ideal gas equation to find out the weight of $ 1 $ mole of gas at the given conditions.
By ideal gas relation we have, $ PV = nRT.....(b) $
Now, we will put the value of volume from (a) in (b)
$ \therefore \,\,\,\,\,\,\,P\dfrac{M}{d} = nRT $
$ \Rightarrow M = \dfrac{{dRT}}{P}......(i) $ (At NTP, $ n = 1 $ )
Now, putting the values of P, V, R, and T in (i), we get;
$ \Rightarrow M = \dfrac{{0.082 \times 0.45 \times 273}}{1} = 10.07 $
The weight of the gas is $ 10g $ .
Now, we will calculate the vapour density of the gas. The Vapour density is given as;
$ Vapour\,Density = \dfrac{{Weight\,of\,given\,gas}}{{Weight\,of\,{H_2}}} $
$ Vapour\,Density = \dfrac{{10}}{2} = 5 $
Hence, The vapour density is $ 5 $ .
So, the molecular weight and vapour density of gas with density $ 0.00045\,g/mL $ are $ 10g $ and $ 5 $ .
Additional Information
The Relative molecular mass is the ratio of the mass of one molecule of a substance to the $ 1/{12^{th}} $ mass of a carbon atom, or $ 1{{ }}amu $ .
Note
Vapour density is the ratio of the weight of a given gas to the weight of hydrogen at a fixed volume measured at the same condition of temperature and pressure. NTP is commonly used as a standard condition for testing and documentation of fan capacities. The volume of $ 1 $ mole of a gas at NTP is $ 22.4L $ .
Formula Used:
Ideal gas law is represented by;
$ PV = nRT $
Where, $ P = $ Pressure of the gas, $ T = $ Temperature of the gas
$ V = $ The volume of the gas, $ n = $ No. of moles of gas and $ R = 0.082 $ (Gas Constant)
$ Vapour\,Density = \dfrac{{Weight\,of\,given\,gas}}{{Weight\,of\,{H_2}}} $
$ density(d) = \dfrac{{mass(M)}}{{volume(V)}} $ or $ volume(V) = \dfrac{{mass(M)}}{{density(d)}}.....(a) $ .
Complete Step-by-step Solution
Given, Density of gas $ = 0.00045\,g/mL = 0.45g/L $
AT NTP condition, $ P = 1\,atm $ , $ V = 1\,L $ , $ R = 0.082 $ and $ T = 273\,K $
Now we will use the ideal gas equation to find out the weight of $ 1 $ mole of gas at the given conditions.
By ideal gas relation we have, $ PV = nRT.....(b) $
Now, we will put the value of volume from (a) in (b)
$ \therefore \,\,\,\,\,\,\,P\dfrac{M}{d} = nRT $
$ \Rightarrow M = \dfrac{{dRT}}{P}......(i) $ (At NTP, $ n = 1 $ )
Now, putting the values of P, V, R, and T in (i), we get;
$ \Rightarrow M = \dfrac{{0.082 \times 0.45 \times 273}}{1} = 10.07 $
The weight of the gas is $ 10g $ .
Now, we will calculate the vapour density of the gas. The Vapour density is given as;
$ Vapour\,Density = \dfrac{{Weight\,of\,given\,gas}}{{Weight\,of\,{H_2}}} $
$ Vapour\,Density = \dfrac{{10}}{2} = 5 $
Hence, The vapour density is $ 5 $ .
So, the molecular weight and vapour density of gas with density $ 0.00045\,g/mL $ are $ 10g $ and $ 5 $ .
Additional Information
The Relative molecular mass is the ratio of the mass of one molecule of a substance to the $ 1/{12^{th}} $ mass of a carbon atom, or $ 1{{ }}amu $ .
Note
Vapour density is the ratio of the weight of a given gas to the weight of hydrogen at a fixed volume measured at the same condition of temperature and pressure. NTP is commonly used as a standard condition for testing and documentation of fan capacities. The volume of $ 1 $ mole of a gas at NTP is $ 22.4L $ .
Recently Updated Pages
Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Class 11 Question and Answer - Your Ultimate Solutions Guide

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

What are Quantum numbers Explain the quantum number class 11 chemistry CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

