Question

Q37) The ratio of partial pressure of SO2 to total pressure of mixture containing 3.2 g of SO2 and 0.18 g water vapours is
a 5/6
b 1/5
c 3/4
d 1/2

Explain Pls​

Answer 1

Answer:

$\alpha log_{8}(2) \geqslant$

Answer 2

ANSWER:

(A) 5/6

Explanation:

Concept Used

The ratio of paprtial pressure is equal to ratio of their respective moles.

i.e

$\frac{Pa}{Pb} = \frac{na}{nb}$

Where,

•Pa and Pb are Partial pressure of Gas a and Gas b respectively

•na and nb are Moles of Gas a and Gas b respectively,

Given In Question

3.2 g of SO2 and

0.18g of H20 (Water)

Now,

Moles of S02= 3.2/64= 0.05

Moles of H2O=0.18/18=0.01

Total moles of mixture= Moles of S02+Moles of H20= 0.05+0.01=0.06

Since,

$\frac{Pa}{Pb} = \frac{na}{nb}$

Similarly,

$\frac{Pa}{Pt} = \frac{na}{nt}$

Here Pt and nt are total pressure and total moles of mixture

Now,

$\frac{P(SO2)}{P(total)}=\frac{n(SO2)}{n(total)}$

$\frac{P(SO2)}{P(total)}=\frac{0.05}{0.06}$

$\frac{P(SO2)}{P(total)}=\frac{5}{6}$

Hence the Right answer is

$\frac{5}{6}$

Hope guys you all are healthy and fit