1 determine the concentration of a solution prepared by...

1). Determine the concentration of a solution prepared by diluting 25.0 mL of a stock 0.188 M Ca(NO3)2 solution to 150.0mL. 2). Determine the molarity of a solution formed by dissolving 468 mg of MgI2 in enough water to yield 50.0 mL of solution. 3). What is the concentration of nitrate ions in a 0.125 M Mg(NO3)2 solution? 4). How many moles of NaCl are required to make 250 mL of a 3.00 M solution? 5). In order to prepare 60 g of 10% Sodium Carbonate solutions, how many grams of water and sodium carbonate is needed? 6). Calculate how may grams of NaCl are needed to prepare a 250 mL of a 0.50 M solution? 7). What volume of 12M Hydrochloric acid must be used to prepare 500 mL of a 6.0 M HCl solution?

Transcript

00:01 Hello students, this question consists of multiple parts.

00:04 In the first part, we have to find out the concentration when the volume is 150 ml.

00:09 The initial concentration and the initial volume has been given.

00:12 We know that the number of moles remains constant throughout the process.

00:18 Therefore, c1v1 will be equal to c2v2.

00:21 We are rearranging this equation in order to get the c2.

00:24 Substituting the value, 0 .188 into 25 divided by 150.

00:31 On solving this, we will get the c2 that is our required concentration will be equal to 0 .0313 m.

00:40 This is the answer to the first question.

00:42 Coming to the second part of the question, here the mass and the volume of the magnesium iodide solution has been given.

00:48 We have to find out the molarity of the solution.

00:51 First of all, we are going to calculate the number of moles which is given by mass divided by molar mass.

00:55 Substituting the value, 468 into 10 to the power minus 3 divided by molar mass is 278 .11.

01:04 On solving this, we will get the number of moles will be equal to 1 .683 into 10 to the power minus 3 moles.

01:12 Substituting this in the molarity formula, that is number of moles divided by volume, 1 .683 into 10 to the power minus 3 divided by 0 .05.

01:23 On solving this, we will get the molarity of the solution will be equal to 33 .66 into 10 to the power minus 3 m.

01:33 This is the answer to the second question.

01:35 In the third part of the question, the concentration of the magnesium nitrate solution has been given.

01:40 We have to find out the concentration of the nitrate ion.

01:42 The magnesium nitrate will dissociate into magnesium ion and 2 nitrate ion.

01:47 Therefore, as it is an ionic component, it dissociate completely.

01:50 Therefore, the concentration of the nitrate ion will be equal to 2 into concentration of the magnesium nitrate ion as it is dissociate into 2 moles of nitrate ion.

02:03 Substituting the value, 2 into 0 .125 m.

02:08 On solving this, we will get the concentration of the nitrate ion will be equal to 0 .25 m.

02:16 This is the answer to the third question.

02:19 The next part of the question, the volume and the concentration of the sodium chloride solution has been given.

02:24 We have to find out the number of moles.

02:25 Number of moles is given by concentration into volume...

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