Question

How would you prepare 1.40 L of a 0.70-M solution of each of the following? from “concentrated” (18 M) sulfuric acid Dilute mL of concentrated to a total volume of 1.40 L with water with 1.40 L water . from “concentrated” (12 M) reagent Dilute mL of concentrated to a total volume of 1.40 L with water with 1.40 L water . from the salt Dissolve g in water, and add water until the total volume of the solution is 1.40 L water is 1.40 L . from “concentrated” (16 M) nitric acid Dilute mL of concentrated to a total volume of 1.40 L with waterwith 1.40 L water . Sodium carbonate from the pure solid Dissolve g in water, and add water until the total volume of the solution is 1.40 Lwater is 1.40 L .

          How would you prepare 1.40 L of a 0.70-M solution of each of the following?
 from “concentrated” (18 M) sulfuric acid
Dilute  mL of concentrated  
  to a total volume of 1.40 L with water  with 1.40 L water
.
 from “concentrated” (12 M) reagent
Dilute  mL of concentrated  
         to a total volume of 1.40 L with water               with 1.40 L water
.
 from the salt
Dissolve  g  in water, and add water until the total volume of the 
         solution is 1.40 L         water is 1.40 L
.
 from “concentrated” (16 M) nitric acid
Dilute  mL of concentrated  
to a total volume of 1.40 L with waterwith 1.40 L water
.
Sodium carbonate from the pure solid
Dissolve  g  in water, and add water until the total volume of the 
solution is 1.40 Lwater is 1.40 L
.

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Added by Joseph R.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

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Solved by Expert Supreeta N

09/06/2023

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From “concentrated” (18 M) sulfuric acid ** Show more…

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How would you prepare 1.40 L of a 0.70-M solution of each of the following? from “concentrated” (18 M) sulfuric acid Dilute mL of concentrated to a total volume of 1.40 L with water with 1.40 L water . from “concentrated” (12 M) reagent Dilute mL of concentrated to a total volume of 1.40 L with water with 1.40 L water . from the salt Dissolve g in water, and add water until the total volume of the solution is 1.40 L water is 1.40 L . from “concentrated” (16 M) nitric acid Dilute mL of concentrated to a total volume of 1.40 L with waterwith 1.40 L water . Sodium carbonate from the pure solid Dissolve g in water, and add water until the total volume of the solution is 1.40 Lwater is 1.40 L .

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Transcript

00:01 Hello students, let's begin with this question.

00:03 Here the molarity is given.

00:05 Let's assume that this molarity is m1 which is equals to 0 .50 molar and the volume is given let's assume that this to be v1 which is equals to 1 .80 liter and here we have to prepare the solution.

00:17 First one is the h2so4 from the concentrated 18 molar h2so4.

00:28 So now here the formula used is m1 v1 is equals to m2 v2.

00:34 So here m1 is 0 .50 molar into 1 .80 liter is equals to m2 is 18 molar into v2.

00:43 So from here this v2 comes to be 0 .05 liter and in terms of ml this v2 comes to be 50 ml.

00:52 Then in the b part here we have to prepare the hcl from the concentrated 12 molar reagent.

01:01 So again here same formula is used that is m1 v1 is equals to m2 v2.

01:07 So from here this m1 is 0 .50 molar into v1 is 1 .80 liter is equals to 12 molar into v2.

01:16 So from here this v2 is equals to 0 .075 liter.

01:21 So from here this v2 comes to be 75 ml.

01:25 Now in the next part here we have to prepare the here c part nicl2 from the salt of nicl2 6h2o.

01:41 So here firstly this will form the nicl2 plus 6h2o.

01:49 So we know the molarity that the molarity is equals to the number of moles of that is dissolved in per liter of the given solution and moles is equals to mass of solute upon the molar mass multiply by here the divided here comes 1 upon volume.

02:11 So from here we can find out the mass...

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