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1 g mixture of CaCO3 and NaCl reacts completely with 100 ml of N/10 HCl. The % of CaCO3 in the mixture is-(1) 40%, (2)60%, (3) 50%, (4) 80%. Also mention which is to be taken as Solute, HCl or CaCO3??

Asked by patra04011965 7th August 2019, 8:57 PM
Answered by Expert
Answer:
Option (3) is correct: 50%
 
Given:
 
Mass of CaCO3 + NaCl = 0.1 gm
 
Volume of HCl = 100 ml
 
Normality of HCl = 0.1 N
 
Molarity of HCl = 0.1 M
 
Molar masss of  CaCo3 = 100 g/mol
 
HCl reacts with CaCo3. The reaction is written as,
 
CaCo3 + 2HCl  → CaCl2 + CO2 +H2O
 
1 mole of CaCo3 reacts with 2 moles of HCl
 
No. of moles of HCl is
 
Molarity equals space fraction numerator No. space of space mole over denominator Volume space in space straight L end fraction

No. space of space moles space equals space 0.1 cross times 0.1

space space space space space space space space space space space space space space space space space space space space space equals 0.01 space mol
 
Now moles of CaCo3 can be calculated as,
 
Moles of  CaCo3 
 
equals fraction numerator 0.01 over denominator 2 end fraction

equals 0.005 space mol
 
 
 
Mass of  CaCo3 is calculated as,
 
No. of moles = fraction numerator Mass over denominator Molar space mass end fraction
 
Mass of  CaCo3 = No. of moles × Molar mass of  CaCo3 
 
                       = 0.005×100
 
                       = 0.5 gm
 
Percentage of CaCo3 = 0.5 × 100
 
                               = 50 %
Answered by Expert 8th August 2019, 11:04 AM
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