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Klio2033 [76]
3 years ago
10

describe how 250 cm³ of 0.2 mol/dm³ H2SO4 could be prepared from 150 cm³ of 1.0mol/dm³ stock solution of the acid​

Chemistry
1 answer:
Reptile [31]3 years ago
5 0

Answer:

Explanation:

250 cm^3 of 0.2 moldm-3 H2SO4 can be prepared from 150cm^3 of 1.0 moldm^-3 by dilution.

150cm^3 of the 1.0 moldm^-3 stock solution is measured out using a measuring cylinder and transferred into a 250 cm^3 standard volumetric flask and made up to mark. The resulting solution is now 250cm^3 of 0.2 moldm-3 H2SO4.

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The answer is:  " 56 g CaCl₂ " .
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Explanation:
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2.0 M CaCl₂  = 2.0 mol CaCl₂ / L  ; 

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                  = moles of solute per L {"Liter"} of solution.
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Given: 250 mL ;   

250 mL = ?  L  ?  ;  


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→ 0.50 mol CaCl₂  .
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1 mol CaCl₂ = ? g ?

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</span>
There are 2 atoms of Cl in " CaCl₂ " ;  

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__________________________________________________________
So, to calculate the molar mass of "CaCl₂" :

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__________________________________________________________
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→  0.50 mol CaCl₂  = ? g CaCl₂  ? ; 

→  0.50 mol CaCl₂ * (111 g CaCl₂ / mol CaCl₂) ;

                                             = (0.50) * (111 g) CaCl₂ ;

                                             =  55.5 g CaCl₂  ;

                                                → round to 2 significant figures; 

                                                →  56 g CaCl₂ .
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The answer is:  " 56 g CaCl₂ " .
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