2 × 10^{-5} {C} charge on {Zn}²+} ions corresponds to

Correct Answer is : 0.64mg0.64 {mg}0.64mg of Zn2+{Zn}²+}Zn2+ ions

2 × 10^{-5} mol of {Zn}²+} ions

1 × 10^{-10} {mol} {Zn}²+} ions

Correct Answer is : 0.64mg0.64 {mg}0.64mg of Zn2+{Zn}²+}Zn2+ ions

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CGPET 2017 Solved Paper

Section: Chemistry

Question : 52 of 150

Marks: +1, -0

2 \times 10^{-5} \mathrm{C}

\mathrm{Zn}^{2+}

ions corresponds to

Solution:

\because 2 \mathrm{F}

amount of charge give

=1 \mathrm{mol}

\mathrm{Zn}^{2+}

ions.

or,

2 \times 96500 \mathrm{C}

amount of charge give

=1

mole of

\mathrm{Zn}^{2+}

ions.

\therefore 2 \times 10^{-5} \mathrm{C}

charge will give

=\frac{2 \times 10^{-5}}{2 \times 96500}

(mole of

\mathrm{Zn}^{2+}

ions)

Amount of

\mathrm{Zn}^{2+}(n)

ions

=1.0 \times 10^{-5}

Also

n=\frac{w}{M}

where

n=

number of moles

w=

mass of

\mathrm{Zn}^{2+}

ions

M=

molar mass of

\mathrm{Zn}^{2+}(65.5)

∴

w=n \times M=1.00 \times 10^{-5} \times 65.5

(M

molar

\mathrm{Zn}^{2+}=65.5)

=65.5 \times 10^{-5} \mathrm{g}

65.5 \times 10^{-5} \times 10^{3}

milligram.

=0.65 \approx 0.64

milligram of

\mathrm{Zn}^{2+}

ions.

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