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Excess of NaOH (aq) was added to 100 mL of FeCl3  (aq) resulting into 2.14 g of Fe(OH)3 . The molarity of FeCl3  (aq) is : (Given molar mass of Fe=56 g mol−1  and molar mass of Cl=35.5 g mol−1 )  
Asked by kmondaldgms | 23 Aug, 2018, 11:04: PM
answered-by-expert Expert Answer
3 NaOH space plus space FeCl subscript 3 subscript open parentheses aq close parentheses end subscript end subscript space rightwards arrow space 3 NaCl subscript open parentheses aq close parentheses end subscript space space plus space straight F open parentheses OH close parentheses subscript 3 subscript open parentheses aq close parentheses end subscript end subscript The space ratio space between space FeCl subscript 3 space and space Fe open parentheses OH close parentheses space is space 1 colon 1 Moles space in space Fe open parentheses OH close parentheses subscript 3 space equals space 56 plus 48 plus 3 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 107 space gm divided by mole Moles space equals fraction numerator Mass over denominator Molar space mass end fraction space space space space space space space space space equals fraction numerator 2.14 over denominator 107 end fraction space space space space space space space space space equals 0.02 space moles Since space the space mole space ratio space is space 1 colon 1 comma space the space moles space of space FeCl subscript 3 space is space 0.02 space moles subscript Molarity space equals moles space pes space litre end subscript 1000 over 100 cross times 0.02 space equals fraction numerator 0.2 over denominator 1000 end fraction space moles Molarity space of space FeCl subscript 3 space is space thus space 0.2 space moles space per space litre
 
 
Molarity of FeCl3 0.2 M
Answered by Varsha | 25 Aug, 2018, 02:39: PM

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