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How do you calculate normality of percent solution? - Answers

Is the makeup of the solution expressed as "percent by mass"? If so, to calculate molarity (or normality), you have to also know the density of the solution Step 1. Lets say the solution is 14%, and the density is 1.09 g/mL. We can write the following: (14 grams solute/100 grams solution) (1.09 grams solution/ mL solution) Step 2. Multiplying and cancelling from step 1 gives you 15.26 grams solute / 100 mL solution. Multiplying top and bottom by 10 gives you 152.6 grams solute per liter. Step 3. Molarity is number of moles per liter. Divide the 152.6 grams of the solute by the forumua weight (or molecular weight) of the solute, and you have the number of moles of solute. This number is therefore the molarity of the solution. If the solution is "percent by volume", the number you have is number of grams per 100 mL. Multiply by 10, and you have grams per liter. Then divide by the formula weight, and you have the molarity.



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How do you calculate normality of percent solution? - Answers

https://math.answers.com/basic-math/How_do_you_calculate_normality_of_percent_solution

Is the makeup of the solution expressed as "percent by mass"? If so, to calculate molarity (or normality), you have to also know the density of the solution Step 1. Lets say the solution is 14%, and the density is 1.09 g/mL. We can write the following: (14 grams solute/100 grams solution) (1.09 grams solution/ mL solution) Step 2. Multiplying and cancelling from step 1 gives you 15.26 grams solute / 100 mL solution. Multiplying top and bottom by 10 gives you 152.6 grams solute per liter. Step 3. Molarity is number of moles per liter. Divide the 152.6 grams of the solute by the forumua weight (or molecular weight) of the solute, and you have the number of moles of solute. This number is therefore the molarity of the solution. If the solution is "percent by volume", the number you have is number of grams per 100 mL. Multiply by 10, and you have grams per liter. Then divide by the formula weight, and you have the molarity.



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https://math.answers.com/basic-math/How_do_you_calculate_normality_of_percent_solution

How do you calculate normality of percent solution? - Answers

Is the makeup of the solution expressed as "percent by mass"? If so, to calculate molarity (or normality), you have to also know the density of the solution Step 1. Lets say the solution is 14%, and the density is 1.09 g/mL. We can write the following: (14 grams solute/100 grams solution) (1.09 grams solution/ mL solution) Step 2. Multiplying and cancelling from step 1 gives you 15.26 grams solute / 100 mL solution. Multiplying top and bottom by 10 gives you 152.6 grams solute per liter. Step 3. Molarity is number of moles per liter. Divide the 152.6 grams of the solute by the forumua weight (or molecular weight) of the solute, and you have the number of moles of solute. This number is therefore the molarity of the solution. If the solution is "percent by volume", the number you have is number of grams per 100 mL. Multiply by 10, and you have grams per liter. Then divide by the formula weight, and you have the molarity.

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      Is the makeup of the solution expressed as "percent by mass"? If so, to calculate molarity (or normality), you have to also know the density of the solution Step 1. Lets say the solution is 14%, and the density is 1.09 g/mL. We can write the following: (14 grams solute/100 grams solution) (1.09 grams solution/ mL solution) Step 2. Multiplying and cancelling from step 1 gives you 15.26 grams solute / 100 mL solution. Multiplying top and bottom by 10 gives you 152.6 grams solute per liter. Step 3. Molarity is number of moles per liter. Divide the 152.6 grams of the solute by the forumua weight (or molecular weight) of the solute, and you have the number of moles of solute. This number is therefore the molarity of the solution. If the solution is "percent by volume", the number you have is number of grams per 100 mL. Multiply by 10, and you have grams per liter. Then divide by the formula weight, and you have the molarity.
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