In a sample of hydrogen sulfide (M = 34.1 g/mol) at a temperature of 3.00 × 102 K, estimate the ratio of the number of molecules that have speeds very close to vrms to the number that have speeds very close to 2vrms.

22.5

Explanation:

Vrms=√((3k_B T)/m)

Let V1 = Vrms and V2 = 2 Vrms

V1=√((3k_B T)/m)  and  V1^2=(3k_B T)/m

V2=2√((3k_B T)/m) V2^2=(12k_B T)/m

Number of molecules is given by dN = N f(v) dV

f(v)=(4 )/п (m/(3k_B T))^(3/2) V^2 e^(-(mv^2)/(2K_B T))

f(v1)=(4 )/п (m/(3k_B T))^(3/2) 〖V1〗^2 e^(-(m〖v1〗^2)/(2K_B T))

f(v2)=(4 )/п (m/(3k_B T))^(3/2) 〖V2〗^2 e^(-(m〖v2〗^2)/(2K_B T))

The ratio of the number of molecules; (dN(v1))/(dN(v2))=Nf(v1)dV/Nf(v2)dV=(f(v1))/(f(v2))

(f(v1))/(f(v2))=((4 )/п (m/(3k_B T))^(3/2) 〖V1〗^2 e^(-(m〖v1〗^2)/(2K_B T)))/((4 )/п (m/(3k_B T))^(3/2) 〖V2〗^2 e^(-(m〖v2〗^2)/(2K_B T)) )=(〖V1〗^2 e^(-(m〖v1〗^2)/(2K_B T)))/(〖V2〗^2 e^(-(m〖v2〗^2)/(2K_B T)) )  

Putting the values of v1 and v2 into  (f(v1))/(f(v2)), the equation above reduces to

(dN(V1))/(dN(V2))=(f(V1))/(f(v2))=1/4  e^(-3/2)/e^(-12/2)  

(dN(V1))/(dN(V2))=22.5

hydrocarbons are organic compounds that contain only hydrogens and carbons. they can be saturated (those which contain only carbon-carbon single bonds) or unsaturated (those which contain carbon-carbon double or triple bonds) in nature. hydrocarbons can undergo several reactions like substitution, elimination etc.  

on substituting one or more hydrogen atoms in hydrocarbon it results in the formation of haloalkane. as the halogen atoms are large compared to the carbon and hydrogen atoms, due to which the molecular weight increases and the bond become polar because of presence electronegative halogen atom and results in the increase in boiling point of the haloalkane.

thus, the correct option is (b) that is the boiling point of the new compound increases on substituting a hydrogen atom with a halogen in a hydrocarbon.