A certain quantity of electricity deposits $0.54 g$ of $A g$ from silver nitrate solution. What volume of hydrogen will be liberated by the same quantity of electricity at $27^{\circ} C$ and $750 m m$ of $H g$ pressure?

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Solution

$\frac{M a s s o f H_{2}}{M a s s o f A g} = \frac{E q . m a s s o f H_{2}}{E q . m a s s o f A g}$
Mass of $H_{2} = \frac{1 \times 0.54}{108} = 0.005 g$
Volume of $H_{2} = \frac{0.005 \times 0.0821 \times 300 \times 760}{2 \times 750} (V = \frac{w}{M} \cdot \frac{R T}{P})$
$= 0.06234 l i t r e$ .

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A certain quantity of electricity deposits 0.54 g of Ag from silver nitrate solution. What volume of hydrogen will be liberated by the same quantity of electricity at 27∘C and 750 mm of Hg pressure?

Mass of H2Mass of Ag=Eq.mass of H2Eq.mass of AgMass of H2=1×0.54108=0.005 gVolume of H2=0.005×0.0821×300×7602×750(V=wM⋅RTP)=0.06234 litre.

A certain quantity of electricity deposits $0.54 g$ of $A g$ from silver nitrate solution. What volume of hydrogen will be liberated by the same quantity of electricity at $27^{\circ} C$ and $750 m m$ of $H g$ pressure?

$\frac{M a s s o f H_{2}}{M a s s o f A g} = \frac{E q . m a s s o f H_{2}}{E q . m a s s o f A g}$
Mass of $H_{2} = \frac{1 \times 0.54}{108} = 0.005 g$
Volume of $H_{2} = \frac{0.005 \times 0.0821 \times 300 \times 760}{2 \times 750} (V = \frac{w}{M} \cdot \frac{R T}{P})$
$= 0.06234 l i t r e$ .