Dissolve 25 g of sodium thiosulphate pentahydrate (Na2S2O3•5H2O, 248 g/mol) in 1 liter of freshly boiled distilled water;
- improve its stability by adding 0.1 g of Na2CO3;
- store overnight, and filter through fine grade paper,
- collect and store in a clean, dark glass bottle.
it depends how many liters or mililiters you want for example:You want to prepare 100mL of 0.1M sodium thiosulfate
first we calculated the grams needed:
100 mL*(0.1 mmole/1 ml)*(248.19 mg/1 mmole)*(1 g/ 1000 mg)
which gives you 2.4819 g of sodium thiosulfate
disolve the 2.4819 grams in 70 mL ofdistilledwater, transfer to a volumetric flask of 100 mL and finish filling it withdistilledwater up to the mark, and then shake to homogenize.
Two thing to be aware about:
S2O32- --> S4O62- + 2e-
So 0.025N = 0.025 mole/L reducing electrons = 0.0125 mol/L Na2S2O3
Need of 1.000 L 'thio' solution to make:
200.22 (g/mol)*0.0125 mol/L = 2.503 g/L to be dissolved, stand overnight, filter it till clear, and titrate on standardized bichromate.
weigh 3.1g sodium thio sulphate for 250ml solution
To prepare 0.1N Sodium Thiosulphate simply dissolve 25g of sodium thiosulphate pentahydrate in 1 liter boiled distilled water.
Q + N = 36;.25Q + .05N = 5.20;.25(36-N) +.05N = 5.20;9- .20N = 5.20;3.8 = .20N;N = 19 and Q = 17
This is a simple algebra question:d + n = 31.1d + .05n = 2.65Subsituting in for "d" using "n" we have:d = 31 - n3.1 - .1n + .05n = 2.65-0.05n = -0.45Solve for n:n = 9d + 9 = 31d = 22Thus, there are 9 nickels and 22 dimes.
The "solution" is a pair of numbers that make the statement true when youplug in the numbers in place of 'q' and 'n'.So far, there are an infinite number of them. In fact, if you graph this equation, thegraph is a straight line, and every point on the line is a solution to the equation.If you expect to get one single number for 'q' and one single number for 'n', thatcan't happen until you have another equation to go along with this one. You alwaysneed the same number of equations as the number of 'unknowns' in them.(Here you have two 'unknowns' ... 'q' and 'n' .)
This problem can be solved using a system of equations.If we call the number of nickles n and the number of dimes d, we can write the following two equations:n + d = 42 (The number of nickels and dimes is 42).05n + .1d = 3.35 (5 cents for every nickel plus 10 cents for every dime is $3.35)Then:n + d = 425n + 10d = 335 (Switch to cents instead of dollars, gets rid of decimals)n = 42 - dWe have now just defined n in terms of d. Now every time you see n in the second equation, you can replace it. This is called substitution.5n + 10d = 335 (from above)5(42 - d) + 10d = 335 (substituting for n)Follow through with the algebra, and d (# of dimes) = 25, then there must be 17 nickels.25 dimes, 17 nickels