Molartiy = moles of solute/Liters of solution
200 mL X (1 Liter/10^2)= 0.200 Liters
0.200 Liters x (0.05 M)= .01 moles of NaOH
.01 moles of NaOH x (40 g NaOH/ 1 mole NaOH)= 0.4 g NaOH are required
(ml NaCl )(1M NaCl) = (200ml NaCl )(0.05M NaCl )
= 10 ml
As %age=mass/volume*100, so
2.5=m/2.5*100, or
m=0.025*2.5=0.0625g
The definition of molarity is moles per litre. Therefore 200x5 is 1 litre. 0.5*5 is 2.5 moles. This is per litre. I.e. 2.5 molar.
The needed mass is 35,549 g.
4 moles or 160 g NaOH is required for one litre solution.
The answer is 6,71 g dried KCl.
Concentration of NaOH = 0.025 M = 0.025 Moles per Litre of SolutionVolume of Solution required = 5.00LWe can say therefore that:Number of Moles of NaOH needed to prepare the solution= Concentration of NaOH * Volume of Solution requiredTherefore:Number of Moles of NaOH needed to prepare the solution= 0.025M * 5.00L= 0.125molesFrom this we can say that 0.125 moles of NaOH are needed to prepare a 5.00 L solution with a concentration of 0.025M of NaOH.
Molarity = moles of solute/liters of solution or, for our purposes moles of solute = liters of solution * Molarity moles of AgNO3 = 0,50 liters * 4.0 M = 2.0 moles of AgNO3 needed --------------------------------------
You need 6,9 mL stock solution.
10
25
The needed mass is 35,549 g.
4 moles or 160 g NaOH is required for one litre solution.
The answer is 6,71 g dried KCl.
Concentration of NaOH = 0.025 M = 0.025 Moles per Litre of SolutionVolume of Solution required = 5.00LWe can say therefore that:Number of Moles of NaOH needed to prepare the solution= Concentration of NaOH * Volume of Solution requiredTherefore:Number of Moles of NaOH needed to prepare the solution= 0.025M * 5.00L= 0.125molesFrom this we can say that 0.125 moles of NaOH are needed to prepare a 5.00 L solution with a concentration of 0.025M of NaOH.
If i understand the question correctly, basically you want to make your standard 1/5 the conc of the stock. Therefore, 300ml stock diluted up with solvent
500ml = 500cm3 = 0.5dm3 0.250M = 0.250mol/dm3 number of moles = molarity x volume number of moles = 0.250mol/dm3 x 0.5dm3 = 0.125mol 0.125mol of NaCl is needed to prepare the required solution.
125 ml 500(ml) * 0.05 = 25 25 / 0.20 = 125
Molarity = moles of solute/liters of solution or, for our purposes moles of solute = liters of solution * Molarity moles of AgNO3 = 0,50 liters * 4.0 M = 2.0 moles of AgNO3 needed --------------------------------------
You need 2,4 g NaOH (0,06 moles).