For each of the following samples of ionic substances,...

For each of the following samples of ionic substances, calculate the number of moles and mass of the positive ions present in each sample. a. $4.25 \mathrm{~g}$ of ammonium iodide, $\mathrm{NH}_{4} \mathrm{I}$ b. 6.31 moles of ammonium sulfide, $\left(\mathrm{NH}_{4}\right)_{2} \mathrm{~S}$ c. $9.71 \mathrm{~g}$ of barium phosphide, $\mathrm{Ba}_{3} \mathrm{P}_{2}$ d. 7.63 moles of calcium phosphate, $\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}$

Transcript

00:01 Hello students, we have given some ions and we have to calculate number of ions that each of the compound or salts will give us.

00:09 So first we have ammonium iodide which gives us nh4 positive and i negative anions.

00:17 Now we have to calculate number of moles of each ions in 4 .25 grams of ammonium iodide.

00:25 Now molar mass of ammonium iodide is 145 gram per mole.

00:29 So this tells us that 145 gram of ammonium iodide is equal to 1 mole.

00:40 Now we will calculate number of moles present in 4 .25 grams.

00:45 So 4 .25 grams moles of salt will contain 1 by 145 multiplied by 4 .25 moles.

00:53 So this value will come out to be equal to 0 .029 moles.

01:04 So initially we have 0 .029 moles of ammonium iodide.

01:11 Therefore in the final the ammonium ions will also be equal to 0 .029 moles and iodide ions will also be equal to 0 .029 moles.

01:27 This will be the answer to first question.

01:32 Our second salt is diammonium sulfide and we have been given it 6 .31 moles.

01:38 So we have directly given moles.

01:40 Now this dissociate into two ammonium ions and one sulfide two negative ions.

01:48 Therefore finally the number of ammonium ions will be equal to double of 6 .31.

01:55 This will be the value of nh4 ions moles and sulfide ions will be equal to the amount of moles of salt itself.

02:05 So this will be 6 .31 moles.

02:10 So here ammonium moles will be equal to 12 .62 that will be double to the sulfide moles...

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