$0.24$ g of a volatile substance, upon vaporisation, gives $45$ ml vapour at NTP. What will be the vapour density of the substance?

95.93

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59.8

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95.39

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5.993

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Solution

The correct option is B $59.8$
The condition at STP are :

$P = 1 a t m$

$T = 273.15 K$

$R = 0.0821 L a t m m o l^{- 1} K^{- 1}$

Using the equation, $P V = \frac{m R T}{M}$
$\Rightarrow M = \frac{m R T}{P V}$

$\Rightarrow M = \frac{0.24 \times 0.0821 \times 273.15}{1 \times 0.045}$

$\Rightarrow M = 119.6$

Since $V a p o r D e n s i t y = \frac{M}{2}$

$\Rightarrow V a p o r D e n s i t y = \frac{119.6}{2} = 59.8$

Therefore, the correct option is B.

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0.24g of a volatile substance, upon vaporisation, gives 45 ml vapour at NTP. What will be the vapour density of the substance?

The correct option is B 59.8The condition at STP are :P=1 atmT=273.15 KR=0.0821 L atm mol−1K−1Using the equation, PV=mRTM⇒M=mRTPV⇒M=0.24×0.0821×273.151×0.045⇒M=119.6Since VaporDensity=M2⇒VaporDensity=119.62=59.8Therefore, the correct option is B.

$0.24$ g of a volatile substance, upon vaporisation, gives $45$ ml vapour at NTP. What will be the vapour density of the substance?