Question

the density of thr vapor of a substance at 373 k, and 1.013 ×10^15 pa is 2.55gm/liter calculate its molecular weight​

Answer 1

Answer:

77

Explanation:

Since,

PV = nRT

P= nRT/V

P= dRT/M

M= dRT/P

= 2.55×0.082×373/1.013(atm)

M = 76.99

Answer 2

The molecular weight is 78 grams.

Given:

The density of the vapour of a substance at 373 k, and 1.013 ×10^15 pa is 2.55gm/litre.

To Find:

The molecular weight.

Solution:

To find the molecular weight we will follow the following steps:

As we know,

PV = nRT

Here, p is the pressure in atm and V is the volume in grams per litre.

P = 1.013 ×10^15 pa = 1 atm

R is gas constant = 0.0821 items

T is the temperature in Calvin and n is the number of moles

$= \frac{m}{M}$

M is molecular mass and is the mass of the substance used.

So,

$pv = \frac{mRT}{M}$

$M = \frac{mRT}{pv}$

$\frac{m}{v} = 2.55gram \: per \: litre$

Now,

$M \: = \frac{ 0.0821 \times 2.55 \times 373 }{1} = 78gram$

Henceforth, the molecular weight is 78 grams.