Equal weights -C-2-H-6- and hydrogen are mixed in an empty container -0-o-C- The fraction of total pressure exerted by hydrogen is-1-21-11-1615-16

Question

Toppr Admin · Accepted Answer

Consider x gm be the weight of C2H6 and hydrogen-No- of moles of hydrogen - x-2No- of moles of C2H6 - x-30Mole fraction of hydrogen - x-2-x-2-x-30- - 15-16The pressure exerted by hydrogen - mole fraction of hydrogen - total pressure-xA0- - 15-16 - 1 - 15-16

Equal weights $C_{2} H_{6}$ and hydrogen are mixed in an empty container at $0^{o} C$ . The fraction of total pressure exerted by hydrogen is-
$1 : 2$
$1 : 1$
$1 : 16$
$15 : 16$

Consider x gm be the weight of $C_{2} H_{6}$ and hydrogen.
No. of moles of hydrogen = x/2
No. of moles of $C_{2} H_{6}$ = x/30
Mole fraction of hydrogen =( x/2)/(x/2+x/30) = 15/16
The pressure exerted by hydrogen = mole fraction of hydrogen * total pressure
= 15/16 * 1 = 15/16

Equal weights C2H6 and hydrogen are mixed in an empty container at 0oC. The fraction of total pressure exerted by hydrogen is-1:21:11:1615:16

Consider x gm be the weight of C2H6 and hydrogen.No. of moles of hydrogen = x/2No. of moles of C2H6 = x/30Mole fraction of hydrogen =( x/2)/(x/2+x/30) = 15/16The pressure exerted by hydrogen = mole fraction of hydrogen * total pressure = 15/16 * 1 = 15/16

Equal weights $C_{2} H_{6}$ and hydrogen are mixed in an empty container at $0^{o} C$ . The fraction of total pressure exerted by hydrogen is-
$1 : 2$
$1 : 1$
$1 : 16$
$15 : 16$

Consider x gm be the weight of $C_{2} H_{6}$ and hydrogen.
No. of moles of hydrogen = x/2
No. of moles of $C_{2} H_{6}$ = x/30
Mole fraction of hydrogen =( x/2)/(x/2+x/30) = 15/16
The pressure exerted by hydrogen = mole fraction of hydrogen * total pressure
= 15/16 * 1 = 15/16