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If 460 cm3 of sulphur (IV) oxide can diffuse through porous partition in 30 seconds, how long will a volume of 620 cm3 hydrogen sulphide take to diffuse through the same partition H=1, S=32, O=16?
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+ Diffusion
+ Chemistry
+ Hydrogen
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Mamagroots
If 460 cm3 of sulphur (IV) oxide can diffuse through porous partition in 30 seconds, how long will a volume of 620 cm3 hydrogen sulphide take to diffuse through the same partition H=1, S=32, O=16?
Graham’s law of diffusion gives that:
R1/R2 =sqrt(M2/M1) ———-(1),
where:
R1=average rate of diffusion of gas 1 (SO2)=V1/t1; V1=460cm^3=volume of SO2 diffusing in time t1; t1=30s;
R2=average rate of diffusion of gas 2 (H2S)=V2/t2; V2=620cm^3=volume of H2S diffusing in time t2; t2=time interval required for the H2S gas to diffuse.
M1=molar mass of SO2=(32+2x16)g=64g
M2=molar mass of H2S =(2x1+32)g =34g
Substituting all values known in equation (1), gives:
(460ml/30s)/(620ml/t2)=sqrt(34/64); and this becomes:
t2[46/(3sx620)]=sqrt(34/64),
t2=[3sx620.sqrt(34/64)]/46 =29.5s
Thus the H2S gas of lower molar mass will take 29.5s to diffuse through the same partition.
R1=average rate of diffusion of gas 1 (SO2)=V1/t1; V1=460cm^3=volume of SO2 diffusing in time t1; t1=30s;
R2=average rate of diffusion of gas 2 (H2S)=V2/t2; V2=620cm^3=volume of H2S diffusing in time t2; t2=time interval required for the H2S gas to diffuse.
M1=molar mass of SO2=(32+2x16)g=64g
M2=molar mass of H2S =(2x1+32)g =34g
Substituting all values known in equation (1), gives:
(460ml/30s)/(620ml/t2)=sqrt(34/64); and this becomes:
t2[46/(3sx620)]=sqrt(34/64),
t2=[3sx620.sqrt(34/64)]/46 =29.5s
Thus the H2S gas of lower molar mass will take 29.5s to diffuse through the same partition.
Graham’s law of diffusion gives that:
R1/R2 =sqrt(M2/M1) ———-(1),
where:
R1=average rate of diffusion of gas 1 (SO2)=V1/t1; V1=460cm^3=volume of SO2 diffusing in time t1; t1=30s;
R2=average rate of diffusion of gas 2 (H2S)=V2/t2; V2=620cm^3=volume of H2S diffusing in time t2; t2=time interval required for the H2S gas to diffuse.
M1=molar mass of SO2=(32+2x16)g=64g
M2=molar mass of H2S =(2x1+32)g =34g
Substituting all values known in equation (1), gives:
(460ml/30s)/(620ml/t2)=sqrt(34/64); and this becomes:
t2[46/(3sx620)]=sqrt(34/64),
t2=[3sx620.sqrt(34/64)]/46 =29.5s
Thus the H2S gas of lower molar mass will take 29.5s to diffuse through the same partition.
Graham’s law of diffusion gives that:
R1/R2 =sqrt(M2/M1) ———-(1),
where:
R1=average rate of diffusion of gas 1 (SO2)=V1/t1; V1=460cm^3=volume of SO2 diffusing in time t1; t1=30s;
R2=average rate of diffusion of gas 2 (H2S)=V2/t2; V2=620cm^3=volume of H2S diffusing in time t2; t2=time interval required for the H2S gas to diffuse.
M1=molar mass of SO2=(32+2x16)g=64g
M2=molar mass of H2S =(2x1+32)g =34g
Substituting all values known in equation (1), gives:
(460ml/30s)/(620ml/t2)=sqrt(34/64); and this becomes:
t2[46/(3sx620)]=sqrt(34/64),
t2=[3sx620.sqrt(34/64)]/46 =29.5s
Thus the H2S gas of lower molar mass will take 29.5s to diffuse through the same partition.
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