Please help with this MCQ?

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2 Answers
Dec 16, 2017

D

Explanation:

First of all you need to find the volume that the vapour would occupy at 273 K.

To get the no. of moles we divide this by #sf(22.4color(white)(x)dm^3)# .

Now we know the no. of moles we can use #sf(n=m/M_r)# to get the #sf(M_r)#.

#sf(V_1=0.025color(white)(x)dm^3" "" "V_2=?#

#sf(T_1=373color(white)(x)K" "" " " "T_2=273color(white)(x)K)#

Since the pressure is constant we can say:

#sf(V_1/T_1=V_2/T_2)#

#:.##sf(V_2=(0.025xx273)/(373)color(white)(x)"dm"^3)#

To find how many moles n will occupy this volume we divide by the molar volume #sf(22.4color(white)(x)dm^3)#:

#:.##sf(n=(0.025xx273)/(373xx22.4)=m/M_r)#

#:.##sf(M_r=(0.10xx373xx22.4)/(0.025xx373))#

This corresponds to response D .

Dec 16, 2017

D

Explanation:

from the idel gases equation you have
#(P_1 xx V_1)/(n_1 xx T_1)=(P_2 xx V_2)/(n_2 xx T_2)#
with #n_1= g/M# = n° of mol, 1 are the real conditons, and 2 the standard conditions
#M= (g xx V_2 xx P_2 xx T_1)/(V_1 xx P_1 xx T_2 xx n_2)#
but#P_1 = P_2# and since #1 L = 1 dm^3# we have
#M= (g xx V_2 xx T_1)/(V_1 xx T_1)= (0,10 g xx 373 K xx 22,4 L)/(0,025 L xx 273 K xx 1 mol) = 122,4 g/(mol)#