Question

Please help with this MCQ?

Answer 1

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 .

Answer 2

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)`