Question

if v ml of vapour of substance at NTP weight W g then molecular weight of substance

Answer 1

The molecular weight of substance is $\frac{W\times 24000}{v}$

Explanation:

According to the ideal gas equation:

$PV=nRT$

P = Pressure of the gas = 1 atm ( at NTP)

V= Volume of the gas = v ml = $\frac{v}{1000}L$   ( 1L=1000ml)

n=  moles of gas =$\frac{\text {given mass}}{\text {Molar mass}}=\frac{Wg}{Mg/mol}$

T= Temperature of the gas in kelvin = 293 K ( at NTP)

R= Gas constant

$1atm\times \frac{v}{1000}L=\frac{W}{M}\times 0.0821Latm/Kmol\times 293K$

$\frac{W}{M}\times 24=\frac{v}{1000}$

$M=\frac{W\times 24000}{v}g/mol$

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Answer 2

The molecular weight of substance is $\frac{W\times 24000}{v}$

v

W×24000

Explanation:

According to the ideal gas equation:

PV=nRTPV=nRT

P = Pressure of the gas = 1 atm ( at NTP)

V= Volume of the gas = v ml = $\frac{v}{1000}L </p><p>1000</p><p>v$

L ( 1L=1000ml)

n= moles of gas =$\frac{\text {given mass}}{\text {Molar mass}}=\frac{Wg}{Mg/mol}$

Molar mass

given mass

=

Mg/mol

Wg

T= Temperature of the gas in kelvin = 293 K ( at NTP)

R= Gas constant

$1atm\times \frac{v}{1000}L=\frac{W}{M}\times 0.0821Latm/Kmol\times 293K1atm× </p><p>1000</p><p>v$

L=

M

W

×0.0821Latm/Kmol×293K

$\frac{W}{M}\times 24=\frac{v}{1000} </p><p>M</p><p>W</p><p> </p><p> ×24= </p><p>1000</p><p>v$

M=$\frac{W\times 24000}{v}g/mol</p><p></p><p>M= </p><p>v</p><p>W×24000</p><p> </p><p> g/mol$