
The charge of how many coulombs is required to deposit $ 1 $ gram of sodium metal (Molar mass $ 23\,\dfrac{g}{{mol}} $ ) from sodium ion is
(A) $ 2098 $
(B) $ 96500 $
(C) $ 193000 $
(D) $ 4196 $
Answer
503.7k+ views
Hint :In the above given question we need to first see how much charge is generated when sodium ion becomes sodium metal and then the ratio of moles of electrons to the moles of sodium metal would be equal. Now calculate moles of sodium in one gram of sodium and equate it to the ratio of charge and sodium metal and then calculate the charge.
Complete Step By Step Answer:
In the above given question, we are asked how much charge is required from sodium ions to deposit one gram of sodium metal.
Now first we need to analyse the reaction for the conversion of sodium ion to sodium metal.
$ N{a^ + } + {e^ - } \to Na $
Now from the above given equation we can say that one mole of sodium ion takes in one mole of electrons to form one mole of sodium metal.
Now we are given that we need to deposit one gram of sodium metal, now first let’s calculate the moles of sodium deposited when one of gram of sodium is there
$ moles = \dfrac{{given\,mass}}{{molecular\,mass}} $
$ \Rightarrow moles = \dfrac{1}{{23}} $
Now from the reaction $ N{a^ + } + {e^ - } \to Na $ , we can say that moles of sodium of charge of electrons are same therefore if $ \dfrac{1}{{23}} $ moles of sodium are formed then $ \dfrac{1}{{23}} $ moles of charge are also formed.
Now we need to calculate the charge of $ \dfrac{1}{{23}} $ moles of electrons. Now each electron has a charge of $ 1.6 \times {10^{ - 19}} $ coulomb and there are $ 6.022 \times {10^{23}} $ units of electrons therefore $ \dfrac{1}{{23}} $ moles of electron will have the charge equivalent to
$ ch\arg e = \dfrac{1}{{23}} \times 6.022 \times {10^{23}} \times 1.6 \times {10^{ - 19}} $
$ \Rightarrow ch\arg e = 4196\,C $
Therefore, the charge required to deposit $ 1 $ gram of sodium metal from sodium ion is $ 4196 $ coulomb.
Note :
Sodium metal is used as a heat exchanger in few nuclear reactors and also as a reagent in the chemical industry. The most abundant thing in which sodium is found are sodium salts. The sodium salt is used for various purposes like de-ice roads in winter which lowers the freezing point of water.
Complete Step By Step Answer:
In the above given question, we are asked how much charge is required from sodium ions to deposit one gram of sodium metal.
Now first we need to analyse the reaction for the conversion of sodium ion to sodium metal.
$ N{a^ + } + {e^ - } \to Na $
Now from the above given equation we can say that one mole of sodium ion takes in one mole of electrons to form one mole of sodium metal.
Now we are given that we need to deposit one gram of sodium metal, now first let’s calculate the moles of sodium deposited when one of gram of sodium is there
$ moles = \dfrac{{given\,mass}}{{molecular\,mass}} $
$ \Rightarrow moles = \dfrac{1}{{23}} $
Now from the reaction $ N{a^ + } + {e^ - } \to Na $ , we can say that moles of sodium of charge of electrons are same therefore if $ \dfrac{1}{{23}} $ moles of sodium are formed then $ \dfrac{1}{{23}} $ moles of charge are also formed.
Now we need to calculate the charge of $ \dfrac{1}{{23}} $ moles of electrons. Now each electron has a charge of $ 1.6 \times {10^{ - 19}} $ coulomb and there are $ 6.022 \times {10^{23}} $ units of electrons therefore $ \dfrac{1}{{23}} $ moles of electron will have the charge equivalent to
$ ch\arg e = \dfrac{1}{{23}} \times 6.022 \times {10^{23}} \times 1.6 \times {10^{ - 19}} $
$ \Rightarrow ch\arg e = 4196\,C $
Therefore, the charge required to deposit $ 1 $ gram of sodium metal from sodium ion is $ 4196 $ coulomb.
Note :
Sodium metal is used as a heat exchanger in few nuclear reactors and also as a reagent in the chemical industry. The most abundant thing in which sodium is found are sodium salts. The sodium salt is used for various purposes like de-ice roads in winter which lowers the freezing point of water.
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