Stock FAQs

how would you prepare 2.46 l of a 6.00 m solution from a 9.00 m stock solution?

by Alexis Effertz Published 3 years ago Updated 2 years ago

How would you prepare 1 L of a 1 m solution?

Describe how would you prepare 1 L of a 1 M solution of sodium chloride. The gram formula weight of sodium chloride is 58.44 g/mol. 1 mole of NaCl = 58.44 grams ( the molar mass of Na which is 22.99 g/mol + the molar mass of chlorine which is 35.45 g/mol = 58.44 g/mol). This amount is the placed in a 1 liter volumetric flask is precisely...

How do you calculate the volume of the target solution?

Rounded to three sig figs and expressed in mililiters, the volume will be So, to prepare your target solution, use a 12.5-mL sample of the stock solution and add enough water to make the volume of the total solution equal to 250.0 mL. This is equivalent to diluting the 12.5-mL sample of the stock solution by a dilution factor of 20.

How do you calculate dilution in chemistry?

ChemistrySolutionsDilution Calculations 1Answer Stefan V. Mar 10, 2015 The answer is d) 62.5 mL. So, you know the volume and the concentrationof your target solution, and the concentrationof your stock solution. If you use the formula for dilution calculations, you'll have #C_1V_1 = C_2V_2#, where

How do you dilute a target solution?

So, to prepare your target solution, use a 12.5-mL sample of the stock solution and add enough water to make the volume of the total solution equal to 250.0 mL. This is equivalent to diluting the 12.5-mL sample of the stock solution by a dilution factor of 20.

What is the difference between solution A and B?

A is a solution of trioxonitrate (v) acid , HNO3, of unknown concentration. B is standard solution of sodium hydroxide containing 4.00g per dm3 of solution. 25cm3 portions solution B required an average of 24.00cm3 of solution A for.

What is solution A?

A is a solution of trioxonitrate (v)acid, HNO3 of unknown concentration.B is standard solution of sodium hydroxide containing 4.00g per dm3 of solution. 25 cm3 portion solution B required an average of 24.00cm3 of solution A for

A B C D E F G H I J K L M N O P Q R S T U V W X Y Z 1 2 3 4 5 6 7 8 9