600 mL of 0,9 % sodium chloride: 6 x 0,9 = 5,4 grams NaCl
125 ml 500(ml) * 0.05 = 25 25 / 0.20 = 125
90 ml of dextrose and 4.41 litres of water.
400 mls would require 40g of glucose for a 10% solution and thus 20g for a 5% solution.
You need 12,5 sodium chloride.
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.
In this instance, 50 mol of sodium chloride is needed and molar mass of NaCl is 58.5 g/mol. Hence the mass we need is 29250 g. But this amount of salt could not be dissolved in 500 ml of water, so we cannot prepare this solution practically.
Sodium chloride is needed to precipitate soap from solutions.
The needed mass is 0,0584 g.
150 mL x 40.0 g LiNO3/100 mL solution = 60.g of solute
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.
8.0 g according to my teacher but i dont know the formula
Final Volume is divided by percent of bleach needed. Lets say you need to make 100mL of 5% bleach solution. So, 100/5 equals 20mL of bleach. You will need to add 20mL to 80mL of water.
Weight to volume solution is calculated by the weight of the solute in grams divided by the volume of the solution in milliliters. Assuming that the solvent is water, then then 100 grams of glucose is needed.
It depends on the volume, if we consider 1 liter of the solution 500 mg of sodium chloride is needed.