19.14 gm of organic acid was dissolved in 20 ml of water it required 25 ml of 0.12 N of NaOH for complete neutralization .then what is equivalent weight of organic acid is???
(1) 58.50 (2) 63.80 (3) 70.95 (4) 68.10
Answers
Answered by
13
Total mass of organic acid = 0.1914 g
No of moles of organic acid = 0.1914/M
60 mL of 0.5 M solution of NaOH
No of moles of NaOH = 0.12 x 25 x 10-3 = 3 x 10-3 mol
No of moles of organic acid = No of moles of NaOH
0.1914/M = 3 x 10-3
M = 0.19143 x 10-3 = 63.8 g
No of moles of organic acid = 0.1914/M
60 mL of 0.5 M solution of NaOH
No of moles of NaOH = 0.12 x 25 x 10-3 = 3 x 10-3 mol
No of moles of organic acid = No of moles of NaOH
0.1914/M = 3 x 10-3
M = 0.19143 x 10-3 = 63.8 g
vishal741:
hii friend answer to yahi hai magar, u used which formula which relations kuchh zamaza nahi
60 ml of 0.5 M solution of NaoH kaha se aaya??
kuchh values jo question mai nahi hai vah kaise nikali
please can you tell me deeply as thinking I have no also little knowledge about that?
okk forget all that just tell me no. of moles of NaOH = 0.12 * 25 *10 - 3 in this what is role of 10 and -3 kaha se aaye??
Answered by
5
Hey,, the answer given by him was right..he use the concept of conservation of mass..
the weight of reactant= weight of product..
Use this relation..
the weight of reactant= weight of product..
Use this relation..
can you tell me in no. of moles of NaOH = 0.12* 25 *10-3 isme 10 - 3 kaise aaye, samaz nahi aaya, please tell me,??
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