My question is that for a binary composite catalyst like 0.5 gram of C3N4 i want to add different weight% (3,5,7......) of Ag3PO4 from (0.3 M solution of AgNO3, 0.1 M solution of Na3PO4 ) , how much volume of both of these required to give me 3,5...... wt% for Ag3PO4? i am getting quite confuse in wt% of liquid.
Dear Rehana,
a) if you want to add x% of 0,5g (of C3N4) of any material (e.g. Ag3PO4) to the C3N4 sample you have to provide x/100*0,5g of Ag3P04 to add to your C3N4 sample; ok?
b) the concentration of your solutions, which provide your Ag and PO4 respectively, are given im M = mol/l (mole per liter); so it is worthwhile here to change the g unit of the Ag3PO4 into mol. We have the mol-weight of Ag3PO4 to 418,58g/mol.
So your x/100*0,5g of Ag3PO4 is equal to x/100*0,5g/(418,58g/mol) = x/100*0,5/418,58 mol of Ag3PO4.
c) fortunately you have 'nice' values of the concentrations of your solutions:
i) the 0,3M solution of AgNO3 is equal to 0,1M solution of Ag3; and
ii) the 0,1M solution of Na3PO4 is equal to 0,1M solution of PO4.
Thus you have to take the same amount from each of the solutions.
d) at a last significant step we have to calculate the amout of solution which corresponds to the x/100*0,5/418,58 mol of Ag3 and or PO4. This step is easily done by:
x/100*0,5/418,58 mol / (0,1M of Ag3 or PO4) = x/100*0,5/418,58 mol /(0,1 mol/liter)=
x/10*0,5/418,58 liter = x*0,05/418,58 liter= x*0,05/418,58 *1000ml=x*0,12ml.
e) So please take x*0,12ml of each of the solutions and add them to your C3N4 sample.
Final remark: please check whether the mol-weight of Ag3PO4 is calculated correctly.
Final remark no. 2: if you want to calculate on how much of solution you have to add in order to have the mentioned x'% in the final sample (thus having x'% of Ag3PO4 and (100-x')% of C3N4) please use x'/(100-x') instead of x/100 at the start of the above series of equations.
Thanks a lot Sir for your kind response.Sir i mean if i take 0.5 gram of C3N4 and then want to add different weight % (3,4,5) of Ag3PO4 made from ( 0.3 M solution of AgNO3, 0.1 M solution of Na3PO4) then how much volume both of these will required to give me weight% for Ag3PO4?
Please have a look at line d) and e) of my calculations:
for a x% weight of Ag3Po4 you have to take x*0,12ml from each of the solutions;*);
that means for 3% weight please take 3*0,12ml=0,36ml of each solution , for 5% please take 5*0,12ml=0,6ml of each solution and so on......
The derivation of the number '0,12ml' is presented in the answer above.
Remark: the unit 'ml' is 1ml =1*milli-liter = 1/1000*liter =1cm³.
*) each of the solutions means: x*0,12ml from solution no.1 and x*0,12ml from solution no.2.
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