Question

IN victor Meyers experiment 0.23g Of a volatile solute displaced air which measures 112ml at NTP. calculate the vapour density and molecular weight of substance.

a. 23.01,46.02
b. 25.2,37.7
c. 52.3,44.2
d. 46.02,23.01​

Answer 1

$\huge {\bold {\underline{ \underline{Given : - }}}}$

Mass of volatile solute is 0.23g

Volume of the gas is 112ml at NTP

$\huge{ \bold {\underline {\underline{To \: find : - }}}}$

Vapour density and Molecular weight of the gas

$\huge{ \bold {\underline {\underline{Solution : - }}}}$

We are given that ,

Volume = 112 ml = 0.112 litres

Mass = 0.23 g

Pressure = 1 atm (NTP conditions)

Temperature = 273K (NTP conditions)

R = 0.0821 lit.atm.mol⁻¹.K⁻¹

Let Molecular weight be 'W'

From ideal gas equation ,

$\large \underline{\bold {\boxed{ \bigstar{ | PV = nRT | }}}}$

Where ,

P is pressure of the gas

V is Volume of the gas

n is Number of moles

R is universal gas constant

T is temperature

Number of moles of the given element or gas is given by ,

$\large {\rm {\underline{ \bold {\boxed{{ n = \frac{given \: weight}{molecular \: \: weight } }}}}}}$

$: \implies \: n \: = \frac{0.23}{W} \rightarrow \: eq(1)$

Now from ideal gas equation ,

$: \implies \: (1)(0.112) = \big( \frac{0.23}{W} \big)(0.0821)(273) \\ \\ : \implies \: 0.112(W) = (0.23)(0.0821)(273) \\ \\ : \implies \: W \: = \bigg( \frac{(0.23)(0.0821)(273)}{0.112} \bigg) \\ \\ : \implies \:W = \bigg( \frac{5.155059}{0.112} \bigg) \\ \\ : \implies {\bold {\boxed{ \pink {W \: = 46.02 \: g}}}}$

Relation between Molecular weight (W) and Vapour density(VD) is given by ,

$\large \underline {\bold {\boxed{ \bigstar { | W = 2 \times VD | }}}}$

Where ,

W is Molecular weight

VD is vapour density

$: \implies \: 46.02 = 2 \times VD \\ \\: \implies \: VD = \bigg( \frac{46.02}{2} \bigg) \\ \\ : \implies {\bold {\boxed {\pink{ VD = 23.01}}}}$

$\dagger \rm \: Hence , \: option(A) \: is \: correct$

$\huge{\bold{\underline{\underline{Additional\:Info:-}}}}$

❈ VALUES OF R IN DIFFERENT UNITS :-

R = 0.0821 lit.atm.mol⁻¹K⁻¹

R = 8.314 × 10⁷ erg.mol⁻¹.K⁻¹

R = 1.987 cal.mol⁻¹.K⁻¹

R = 8.314 J.mol⁻¹.K⁻¹