0.829 M is the molarity.
The formula is:
mL x M = millimoles of substance
millimoles of NaOH = millimoles of HCl
mL x M = mL x M
28.6 x 0.145 = 5.00 x M
M = 0.829
1 Na2CO3 / 2 HCl
28.45 mL HCl
1 L
1000 mL
= .02845 L
.1586 g Na2CO3
1 mol
105.98 g
= .0014965088 mol
.0014965088 mol
.02845 L
= .0526013629
.0526013629 x 2 = .1052027259 = .1052 M
28.6ml/10ml*.175M = .5005 M then with significant figured the answer is 0.501 M
The molarity is 0.496m.
(100ml)(0.125M NaOH) = (500ml)(X Molarity) Molarity = 0.025 M
There would be 117g of NaCl in 1000 ml or 1 liter. The MWt of NaCl is 58.5 so 117/58.5 is 2.00 - so the molarity is 2.00 moles per liter of NaCl
It means that 100mL of solution has 75g of solute dissolved in it.
100ppm means, you need 100mg in 1oooml. Thus, for 100ml solution, you will need 10mg of methanol to make 100ppm solution.
There isn't enough to answer your question. You've given a volume, but haven't given a weight or amount of moles. I also do not know if there is a relevance to your including the graduated cylinder. A solution should always be made up in a volumetric flask.
1
(100ml)(0.125M NaOH) = (500ml)(X Molarity) Molarity = 0.025 M
There would be 117g of NaCl in 1000 ml or 1 liter. The MWt of NaCl is 58.5 so 117/58.5 is 2.00 - so the molarity is 2.00 moles per liter of NaCl
It means that 100mL of solution has 75g of solute dissolved in it.
Molarity is the number of moles of solute dissolved in one liter of solution. The units of molarity are moles per litre iemoles of solute per litre of solution. The equation for molarity is: (moles of solute)/(litres of solution) Eg 3 moles per litre or 3 mol L-1 Alternatively, volume can be measured in cubic decimeters: dm3 which are equal to litres. Eg 3 moles dm-3 is the same as 3 mol L-1 See:http://dl.clackamas.edu/ch105-04/molarity.htm
100ppm means, you need 100mg in 1oooml. Thus, for 100ml solution, you will need 10mg of methanol to make 100ppm solution.
The molarity remain unchanged, only the amount of NaOH is changed.
It would be 12.6g of IKI to obtain the 100mL solution of 0.300 M IKI.
It would be 12.6g of IKI to obtain the 100mL solution of 0.300 M IKI.
58 grams of NaCl in cylinder measure water to 100ml
0.01
Iodine solution is an efficient indicator for starch. It will go from a light brown colour to a black colour if starch is present and will stay light brown if no starch is present. Hope this Helps!