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Len [333]
2 years ago
12

Find the molarity of a 500 L solution that contains 10 moles of fluorine.

Chemistry
1 answer:
deff fn [24]2 years ago
3 0

<u>Given </u><u>:</u><u>-</u>

  • Number of moles = 10
  • Volume of solution = 500L

<u>To </u><u>Find</u><u> </u><u>:</u><u>-</u>

  • Molarity of the solution .

<u>Solution</u><u> </u><u>:</u><u>-</u>

As we know that ,

\longrightarrow Molarity (M )=\dfrac{Number\ of \ moles \ of \ solute }{Volume\ of \ solution\ (in \ L) }

Substitute ,

\longrightarrow M =\dfrac{10}{500L}

Simplify,

\longrightarrow M = \dfrac{1}{50}

Convert into decimal ,

\longrightarrow \underline{\underline{ M = 0.02 \ mol \ L^{-1}}}

This is the required answer .

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If 34.7 g of AgNO₃ react with 28.6 g of H₂SO₄ according to this UNBALANCED equation below, how many grams of Ag₂SO₄ could be for
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The  number  of grams   of Ag2SO4  that could be formed  is   31.8  grams



    <u><em> calculation</em></u>

Balanced   equation is  as below

2 AgNO3 (aq)  + H2SO4(aq)  →  Ag2SO4 (s)   +2 HNO3 (aq)


  • Find  the  moles  of  each reactant by use  of  mole= mass/molar mass  formula

that is  moles of  AgNO3= 34.7 g / 169.87  g/mol= 0.204 moles

             moles of  H2SO4 =  28.6  g/98  g/mol  =0.292  moles

  • use the  mole  ratio to determine the moles of  Ag2SO4

   that is;

  •    the mole ratio of  AgNo3 : Ag2SO4 is  2:1 therefore  the  moles of Ag2SO4=  0.204  x1/2=0.102 moles

  • The moles  ratio of H2SO4  : Ag2SO4  is  1:1  therefore  the moles of Ag2SO4 = 0.292  moles

 

  •      AgNO3  is the limiting reagent therefore  the moles of   Ag2SO4 = 0.102  moles

<h3>     finally  find  the mass  of Ag2SO4  by use of    mass=mole  x molar mass  formula</h3>

that  is  0.102   moles  x  311.8  g/mol= 31.8 grams

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