HomeLearn Fast select Course
Show Unit/ChapterChange UnitChange SubjectLast Menu
NEET-XII-Physics

33: Thermal and Chemical Effects of Electric Current

with Solutions - page 5
Qstn# iv-20 Prvs-QstnNext-Qstn
  • #20
    Two voltameters, one with a solution of silver salt and the other with a trivalent-metal salt, are connected in series and a current of 2 A is maintained for 1.50 hours. It is found that 1.00 g of the trivalent metal is deposited. (a) What is the atomic weight of the trivalent metal? (b) How much silver is deposited during this period? Atomic weight of silver is 107.9 g mol-1. (a) What is the atomic weight of the trivalent metal? (b) How much silver is deposited during this period? Atomic weight of silver is 107.9 g mol-1.
    Ans : Given:
    Mass of salt deposited, m = 1 g
    Current, i = 2 A
    Time, t = 1.5 hours = 5400 s
    For the trivalent metal salt:
    Equivalent mass = `` \frac{1}{3}``Atomic weight
    The E.C.E of the salt,
    `` Z=\frac{\,\mathrm{\,Equivalent\,}\,\mathrm{\,mass\,}}{96500}=\frac{\,\mathrm{\,Atomic\,}\,\mathrm{\,weight\,}}{3\times 96500}`` (a) Using the formula, m = Zit, we get:
    `` 1\times {10}^{-3}=\frac{\,\mathrm{\,Atomic\,}\,\mathrm{\,weight\,}}{3\times 96500}\times 2\times 5400``
    `` \Rightarrow \,\mathrm{\,Atomic\,}\,\mathrm{\,weight\,}=\frac{3\times 96500\times {10}^{-3}}{2\times 5400}=26.8\times {10}^{-3}\,\mathrm{\,kg\,}/\,\mathrm{\,mole\,}``
    `` \Rightarrow \,\mathrm{\,Atomic\,}\,\mathrm{\,weight\,}=26.8\,\mathrm{\,g\,}/\,\mathrm{\,mole\,}`` (b) Using the relation between equivalent mass and mass deposited on plates, we get:
    `` \frac{{E}_{\mathit{1}}}{{E}_{\mathit{2}}}=\frac{{m}_{1}}{{m}_{2}}``
    `` \Rightarrow \frac{26.8}{3\times 107.9}=\frac{1}{{m}_{2}}``
    `` \Rightarrow {m}_{2}=12.1\,\mathrm{\,g\,}``
    `` ``
    Page No 219: (a) Using the formula, m = Zit, we get:
    `` 1\times {10}^{-3}=\frac{\,\mathrm{\,Atomic\,}\,\mathrm{\,weight\,}}{3\times 96500}\times 2\times 5400``
    `` \Rightarrow \,\mathrm{\,Atomic\,}\,\mathrm{\,weight\,}=\frac{3\times 96500\times {10}^{-3}}{2\times 5400}=26.8\times {10}^{-3}\,\mathrm{\,kg\,}/\,\mathrm{\,mole\,}``
    `` \Rightarrow \,\mathrm{\,Atomic\,}\,\mathrm{\,weight\,}=26.8\,\mathrm{\,g\,}/\,\mathrm{\,mole\,}`` (b) Using the relation between equivalent mass and mass deposited on plates, we get:
    `` \frac{{E}_{\mathit{1}}}{{E}_{\mathit{2}}}=\frac{{m}_{1}}{{m}_{2}}``
    `` \Rightarrow \frac{26.8}{3\times 107.9}=\frac{1}{{m}_{2}}``
    `` \Rightarrow {m}_{2}=12.1\,\mathrm{\,g\,}``
    `` ``
    Page No 219:
  • #20-a
    What is the atomic weight of the trivalent metal?
    Ans : Using the formula, m = Zit, we get:
    `` 1\times {10}^{-3}=\frac{\,\mathrm{\,Atomic\,}\,\mathrm{\,weight\,}}{3\times 96500}\times 2\times 5400``
    `` \Rightarrow \,\mathrm{\,Atomic\,}\,\mathrm{\,weight\,}=\frac{3\times 96500\times {10}^{-3}}{2\times 5400}=26.8\times {10}^{-3}\,\mathrm{\,kg\,}/\,\mathrm{\,mole\,}``
    `` \Rightarrow \,\mathrm{\,Atomic\,}\,\mathrm{\,weight\,}=26.8\,\mathrm{\,g\,}/\,\mathrm{\,mole\,}``
  • #20-b
    How much silver is deposited during this period? Atomic weight of silver is 107.9 g mol-1.
    Ans : Using the relation between equivalent mass and mass deposited on plates, we get:
    `` \frac{{E}_{\mathit{1}}}{{E}_{\mathit{2}}}=\frac{{m}_{1}}{{m}_{2}}``
    `` \Rightarrow \frac{26.8}{3\times 107.9}=\frac{1}{{m}_{2}}``
    `` \Rightarrow {m}_{2}=12.1\,\mathrm{\,g\,}``
    `` ``
    Page No 219: