$1 m m H g$ represents a pressure of :

760 a t m

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\frac{1}{760} a t m

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\frac{1}{76} a t m

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\frac{1}{76} \times 10^{- 2} a t m

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Solution

The correct option is B $\frac{1}{760} a t m$
1 $m m$ of $H g$ is a manometric unit of pressure formerly defined as the extra pressure generated by a column of mercury one $m m$ high. $1 m m$ of $H g$ is approx $1 t o r r$ or $1 / 760$ of standard atmospheric pressure.

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Similar questions

Q. Following reaction occurs at

25^{o} C

2 N O (g, 1 \times 10^{- 5} a t m) + {C l}_{2} (g, 1 \times 10^{- 2} a t m) ⇌ 2 N O C l (g, 1 \times 10^{- 2} a t m)

Δ G^{o}

is_______________.

Q. Following reaction occurs at

25^{\circ} C

2 N O (g, 1 \times 10^{- 5} a t m) + C l_{2} (g, 1 \times 10^{- 2} a t m) ⇌ 2 N O C l (g, 1 \times 10^{- 2} a t m)

The value of

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When 1.0 mole of $N_{2}$ and 3.0 moles of $H_{2}$ was heated in a vessel at 773K and a pressure of 3.55 atm. 30% of $N_{2}$ is converted into $N H_{3}$ at equilibrium. Calculate KP for the reaction -

The vapour pressure of dilute aqueous solution of glucose is 750 mm of Hg at 373 K. The mole fraction of glucose is :

76 c m

H g

pressure is equivalent to

76

pascal.

True =1; False=0

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1mmHg represents a pressure of :

The correct option is B 1760atm1 mm of Hg is a manometric unit of pressure formerly defined as the extra pressure generated by a column of mercury one mm high. 1mm of Hg is approx 1torr or 1/760 of standard atmospheric pressure.

$1 m m H g$ represents a pressure of :

The correct option is B $\frac{1}{760} a t m$
1 $m m$ of $H g$ is a manometric unit of pressure formerly defined as the extra pressure generated by a column of mercury one $m m$ high. $1 m m$ of $H g$ is approx $1 t o r r$ or $1 / 760$ of standard atmospheric pressure.