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Berapakah jumlah mol 700 gram batu kapur (CaCO3) jika diketahui Ar: Ca = 40, C = 12, dan O = 16?

Pertanyaan

Berapakah jumlah mol 700 gram batu kapur (CaCO3) jika diketahui Ar: Ca = 40, C = 12, dan O = 16?

Pembahasan Soal:

menentukan Mr CaCO3 :

M subscript r space end subscript Ca C O subscript 3 double bond A subscript r space Ca and A subscript r space C plus left parenthesis 3 cross times A subscript r space O right parenthesis space space space space space space space space space space space space equals 40 plus 12 plus left parenthesis 3 cross times 16 right parenthesis space space space space space space space space space space space space equals 100 

menentukan mol CaCO3 :

mol space equals massa over M subscript r mol space equals 700 over 100 mol space equals 7 space mol 

Jadi, mol dari CaCOadalah 7 mol.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

A. Nurul

Mahasiswa/Alumni UIN Syarif Hidayatullah Jakarta

Terakhir diupdate 04 Juni 2021

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Pertanyaan yang serupa

Tentukan Mr dari senyawa berikut. a.   b.

Pembahasan Soal:

Massa molekul adalah penambahan dari massa atom-atom penyusun suatu molekul

Pada senyawa begin mathsize 14px style Al subscript 2 open parentheses S O subscript 4 close parentheses subscript 3 end style terdiri dari 2 Al, 3 S dan 12 O dengan massa atom Al=27, S=32 dan O=16 maka massa molekulnya


begin mathsize 14px style left parenthesis 2 cross times 27 right parenthesis plus left parenthesis 3 cross times 32 right parenthesis plus left parenthesis 12 cross times 16 right parenthesis equals 342 end style 


Pada senyawa begin mathsize 14px style Cu S O subscript 4 point 5 H subscript 2 O end style terdiri dari 1 Cu, 1 S, 4 O dan 5 molekul air dengan massa atom Cu=63,5, S=32, O=16 dan H=1 maka massa molekulnya


begin mathsize 14px style 63 comma 5 plus 32 plus left parenthesis 16 cross times 4 right parenthesis plus left parenthesis 5 cross times left parenthesis left parenthesis 2 cross times 1 right parenthesis plus 16 right parenthesis right parenthesis equals 294 comma 5 end style


Jadi, massa molekul a adalah 315 dan massa molekul b adalah 294,5.space 

0

Roboguru

Lengkapi tabel berikut ini!

Pembahasan Soal:

Langkah 1: Melengkapi tabel NO

1. Mol NO (n)

    table attributes columnalign right center left columnspacing 0px end attributes row cell n subscript NO end cell equals cell fraction numerator jumlah space partikel subscript NO over denominator L end fraction end cell row blank equals cell fraction numerator 6 comma 02 cross times 10 to the power of 23 space over denominator 6 comma 022 cross times 10 to the power of 23 end fraction end cell row blank equals cell 1 space mol end cell end table


2. Mr NO 

    table attributes columnalign right center left columnspacing 0px end attributes row cell Mr subscript NO end cell equals cell Ar subscript N and Ar subscript O end cell row blank equals cell 14 plus 16 end cell row blank equals cell 30 space begin inline style bevelled gram over mol end style end cell end table


3. Massa NO (m)

    table attributes columnalign right center left columnspacing 0px end attributes row cell m subscript NO end cell equals cell n subscript NO cross times Mr subscript NO end cell row blank equals cell 1 space mol space cross times 30 space begin inline style bevelled gram over mol end style end cell row blank equals cell 30 space gram end cell end table


4. Volume NO (STP)

    table attributes columnalign right center left columnspacing 0px end attributes row cell V subscript NO end cell equals cell n subscript NO cross times 22 comma 4 space begin inline style bevelled L over mol end style end cell row blank equals cell 1 space mol space cross times 22 comma 4 space begin inline style bevelled L over mol end style end cell row blank equals cell 22 comma 4 space L end cell end table


5. Volume NO (27bold degree bold space C, 1 atm)

    table attributes columnalign right center left columnspacing 0px end attributes row cell V subscript NO end cell equals cell n subscript NO cross times 24 space begin inline style bevelled L over mol end style end cell row blank equals cell 1 space mol cross times 24 space begin inline style bevelled L over mol end style end cell row blank equals cell 24 space L end cell end table


Langkah 2: Melangkapi tabel H subscript bold 2 

1. Jumlah partikel H subscript bold 2 (x)

    table attributes columnalign right center left columnspacing 0px end attributes row cell Jumlah space partikel subscript H subscript 2 end subscript end cell equals cell n subscript H subscript 2 end subscript cross times L end cell row blank equals cell 0 comma 1 space mol cross times 6 comma 022 cross times 10 to the power of 23 space partikel end cell row blank equals cell 6 comma 022 cross times 10 to the power of 22 space partikel end cell row blank blank blank end table

2. Mr H subscript bold 2

    table attributes columnalign right center left columnspacing 0px end attributes row cell Mr subscript H subscript 2 end subscript end cell equals cell 2 cross times Ar subscript H end cell row blank equals cell 2 cross times 1 space begin inline style bevelled gram over mol end style end cell row blank equals cell 2 space begin inline style bevelled gram over mol end style end cell end table
 

3. Massa H subscript bold 2

    table attributes columnalign right center left columnspacing 0px end attributes row cell m subscript H subscript 2 end subscript end cell equals cell n subscript H subscript 2 end subscript cross times Mr subscript H subscript 2 end subscript end cell row blank equals cell 0 comma 1 space mol space cross times 2 space begin inline style bevelled gram over mol end style end cell row blank equals cell 0 comma 2 space gram end cell end table


4. Volume H subscript bold 2 (STP)

    table attributes columnalign right center left columnspacing 0px end attributes row cell V subscript H subscript 2 end subscript end cell equals cell n subscript H subscript 2 end subscript cross times 22 comma 4 space begin inline style bevelled L over mol end style end cell row blank equals cell 0 comma 1 space mol space cross times 22 comma 4 space begin inline style bevelled L over mol end style end cell row blank equals cell 2 comma 24 space L end cell end table


5. Volume H subscript bold 2 (27bold degree bold space C, 1 atm)

    table attributes columnalign right center left columnspacing 0px end attributes row cell V subscript H subscript 2 end subscript end cell equals cell n subscript H subscript 2 end subscript cross times 24 space begin inline style bevelled L over mol end style end cell row blank equals cell 0 comma 1 space mol cross times 24 space begin inline style bevelled L over mol end style end cell row blank equals cell 2 comma 4 space L end cell end table


Langkah 3: Melangkapi tabel N H subscript bold 3

1. mol N H subscript bold 3 (n)

    table attributes columnalign right center left columnspacing 0px end attributes row cell n subscript N H subscript 3 end subscript end cell equals cell fraction numerator V subscript STP subscript N H subscript 3 end subscript end subscript over denominator 22 comma 4 space bevelled L over mol end fraction end cell row blank equals cell fraction numerator 4 comma 48 space L over denominator 22 comma 4 space bevelled L over mol end fraction end cell row blank equals cell 0 comma 2 space mol end cell end table
 

2. Jumlah partikel N H subscript bold 3 (x)
     table attributes columnalign right center left columnspacing 0px end attributes row cell Jumlah space partikel subscript N H subscript 3 end subscript end cell equals cell n subscript N H subscript 3 end subscript cross times L end cell row blank equals cell 0 comma 2 space mol cross times 6 comma 022 cross times 10 to the power of 23 space partikel end cell row blank equals cell 1 comma 2 cross times 10 to the power of 23 space partikel end cell end table


3. Mr N H subscript bold 3

    table attributes columnalign right center left columnspacing 0px end attributes row cell Mr subscript N H subscript 3 end subscript end cell equals cell Ar subscript N plus left parenthesis 3 cross times Ar subscript H right parenthesis end cell row blank equals cell 14 space begin inline style bevelled gram over mol end style plus left parenthesis 3 cross times 1 space begin inline style bevelled gram over mol right parenthesis end style end cell row blank equals cell 17 space begin inline style bevelled gram over mol end style end cell end table
 

4. Massa N H subscript bold 3

    table attributes columnalign right center left columnspacing 0px end attributes row cell m subscript N H subscript 3 end subscript end cell equals cell n subscript N H subscript 3 end subscript cross times Mr subscript N H subscript 3 end subscript end cell row blank equals cell 0 comma 2 space mol space cross times 17 space begin inline style bevelled gram over mol end style end cell row blank equals cell 3 comma 4 space gram end cell end table


5. Volume N H subscript bold 3 (27bold degree bold space C, 1 atm)

    table attributes columnalign right center left columnspacing 0px end attributes row cell V subscript N H subscript 3 end subscript end cell equals cell n subscript N H subscript 3 end subscript cross times 24 space begin inline style bevelled L over mol end style end cell row blank equals cell 0 comma 2 space mol cross times 24 space begin inline style bevelled L over mol end style end cell row blank equals cell 4 comma 8 space L end cell end table


Langkah 4: Melangkapi tabel C H subscript 4

1. mol C H subscript 4 (n)

    table attributes columnalign right center left columnspacing 0px end attributes row cell n subscript C H subscript 4 end subscript end cell equals cell fraction numerator V subscript RTP subscript C H subscript 4 end subscript end subscript over denominator 24 space bevelled L over mol end fraction end cell row blank equals cell fraction numerator 12 comma 3 space L over denominator 24 space bevelled L over mol end fraction end cell row blank equals cell 0 comma 5125 space mol end cell end table 
 

2. Jumlah partikel C H subscript 4 (x)
     table attributes columnalign right center left columnspacing 0px end attributes row cell Jumlah space partikel subscript C H subscript 4 end subscript end cell equals cell n subscript C H subscript 4 end subscript cross times L end cell row blank equals cell 0 comma 512 space mol cross times 6 comma 022 cross times 10 to the power of 23 space partikel end cell row blank equals cell 3 comma 08 cross times 10 to the power of 23 space partikel end cell end table


3. Mr C H subscript 4

    table attributes columnalign right center left columnspacing 0px end attributes row cell Mr subscript C H subscript 4 end subscript end cell equals cell Ar subscript C plus left parenthesis 4 cross times Ar subscript H right parenthesis end cell row blank equals cell 12 space begin inline style bevelled gram over mol end style plus left parenthesis 4 cross times 1 space begin inline style bevelled gram over mol right parenthesis end style end cell row blank equals cell 16 space begin inline style bevelled gram over mol end style end cell end table
 

4. Massa C H subscript 4

    table attributes columnalign right center left columnspacing 0px end attributes row cell m subscript C H subscript 4 end subscript end cell equals cell n subscript C H subscript 4 end subscript cross times Mr subscript C H subscript 4 end subscript end cell row blank equals cell 0 comma 512 space mol space cross times 16 space begin inline style bevelled gram over mol end style end cell row blank equals cell 8 comma 2 space gram end cell end table 


5. Volume C H subscript 4 (STP)

    table attributes columnalign right center left columnspacing 0px end attributes row cell V subscript C H subscript 4 end subscript end cell equals cell n subscript C H subscript 4 end subscript cross times 22 comma 4 space begin inline style bevelled L over mol end style end cell row blank equals cell 0 comma 512 space mol cross times 22.4 space begin inline style bevelled L over mol end style end cell row blank equals cell 11 comma 48 space L end cell end table  


Dengan demikian, tabel lengkapnya adalah

0

Roboguru

Lengkapi tabel berikut!

Pembahasan Soal:

Mol begin mathsize 14px style N H subscript 3 end style

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row mol equals cell massa over M subscript r end cell row blank equals cell fraction numerator 34 space g over denominator 17 space g forward slash mol end fraction end cell row blank equals cell 2 space mol end cell end table end style 

Volume STP begin mathsize 14px style N H subscript 3 end style

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell V subscript STP space end cell equals cell mol cross times 22 comma 4 end cell row blank equals cell 2 cross times 22 comma 4 end cell row blank equals cell 44 comma 8 space L end cell end table end style  

Jumlah Partikel begin mathsize 14px style N H subscript 3 end style

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell Jumlah space partikel end cell equals cell mol cross times L end cell row blank equals cell 2 cross times 6 comma 02 cross times 10 to the power of 23 end cell row blank equals cell 12 comma 04 cross times 10 to the power of 23 end cell row blank equals cell 1 comma 204 cross times 10 to the power of 24 end cell end table end style  

Massa begin mathsize 14px style H subscript 2 O end style 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row mol equals cell massa over M subscript r end cell row cell 0 comma 1 end cell equals cell fraction numerator massa over denominator 18 space g forward slash mol end fraction end cell row massa equals cell 1 comma 8 space g end cell end table end style 

Volume STP begin mathsize 14px style H subscript 2 O end style

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell V subscript STP space end cell equals cell mol cross times 22 comma 4 end cell row blank equals cell 0 comma 1 cross times 22 comma 4 end cell row blank equals cell 2 comma 24 space L end cell end table end style 

Jumlah partikel begin mathsize 14px style H subscript 2 O end style

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell Jumlah space partikel end cell equals cell mol cross times L end cell row blank equals cell 0 comma 1 cross times 6 comma 02 cross times 10 to the power of 23 end cell row blank equals cell 6 comma 02 cross times 10 to the power of 22 end cell end table end style  

Mol begin mathsize 14px style C O subscript 2 end style 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell V subscript STP space end cell equals cell mol cross times 22 comma 4 end cell row cell 5 comma 6 space L end cell equals cell mol cross times 22 comma 4 end cell row mol equals cell 0 comma 25 space mol end cell end table end style 

Massa begin mathsize 14px style C O subscript 2 end style

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row mol equals cell massa over M subscript r end cell row cell 0 comma 25 end cell equals cell fraction numerator massa over denominator 44 space g forward slash mol end fraction end cell row massa equals cell 11 space g end cell end table end style 

Jumlah partikel begin mathsize 14px style C O subscript 2 end style

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell Jumlah space partikel end cell equals cell mol cross times L end cell row blank equals cell 0 comma 25 cross times 6 comma 02 cross times 10 to the power of 23 end cell row blank equals cell 1 comma 505 cross times 10 to the power of 23 end cell end table end style 

Mol begin mathsize 14px style O subscript 2 end style 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell Jumlah space partikel end cell equals cell mol cross times L end cell row cell 6 comma 02 cross times 10 blank to the power of 23 end cell equals cell mol cross times 6 comma 02 cross times 10 to the power of 23 end cell row mol equals cell 1 space mol end cell end table end style 

Massa begin mathsize 14px style O subscript 2 end style

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row mol equals cell massa over M subscript r end cell row 1 equals cell fraction numerator massa over denominator 32 space g forward slash mol end fraction end cell row massa equals cell 32 space g end cell end table end style 

Volume STP begin mathsize 14px style O subscript 2 end style

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell V subscript STP space end cell equals cell mol cross times 22 comma 4 end cell row blank equals cell 1 cross times 22 comma 4 end cell row blank equals cell 22 comma 4 space L end cell end table end style 

Diisikan ke tabel:

 

Jadi, tabel lengkapnya adalah seprti diatas.

0

Roboguru

Hitunglah dari  (diketahui data Massa Atom Relatif : P = 31, Fe = 56, S = 32, K = 39, Al = 27, H = 1, O = 16)

Pembahasan Soal:

Massa molekul relatif suatu senyawa molekul merupakan jumlah massa atom relatif dari seluruh atom penyusun senyawa tersebut.


table attributes columnalign right center left columnspacing 0px end attributes row cell M subscript r space Fe subscript 2 open parentheses S O subscript 4 close parentheses subscript 3 end cell equals cell open parentheses 2 cross times A subscript r space Fe close parentheses plus open parentheses 3 cross times A subscript r space S close parentheses plus open parentheses 12 cross times A subscript r space O close parentheses end cell row cell M subscript r space Fe subscript 2 open parentheses S O subscript 4 close parentheses subscript 3 end cell equals cell open parentheses 2 cross times 56 close parentheses plus open parentheses 3 cross times 32 close parentheses plus open parentheses 12 cross times 16 close parentheses end cell row cell M subscript r space Fe subscript 2 open parentheses S O subscript 4 close parentheses subscript 3 end cell equals cell 112 plus 96 plus 192 end cell row cell italic M subscript italic r bold space Fe subscript bold 2 begin bold style open parentheses S O subscript 4 close parentheses end style subscript bold 3 end cell bold equals cell bold 400 bold space italic g bold forward slash bold mol end cell end table


Jadi, italic M subscript italic r bold space Fe subscript bold 2 begin bold style open parentheses S O subscript 4 close parentheses end style subscript bold 3 adalah 400 g/mol.space

0

Roboguru

Massa 0,25 mol kalsium karbonat  (25 gr) (Ar C= 12, 0=16, Ca = 40).

Pembahasan Soal:

Langkah 1: Menghitung Mr Ca C O subscript bold 3

table attributes columnalign right center left columnspacing 0px end attributes row cell Mr subscript Ca C O subscript 3 end subscript end cell equals cell Ar space Ca and Ar space C plus left parenthesis 3 cross times Ar space O right parenthesis end cell row blank equals cell 40 plus 12 plus left parenthesis 3 cross times 16 right parenthesis end cell row blank equals cell 40 plus 12 plus 48 end cell row blank equals cell 100 space begin inline style bevelled gram over mol end style end cell end table


Langkah 2: Menghitung massa Ca C O subscript bold 3 

table attributes columnalign right center left columnspacing 0px end attributes row cell n subscript Ca C O subscript 3 end subscript end cell equals cell m subscript Ca C O subscript 3 end subscript over Mr subscript Ca C O subscript 3 end subscript end cell row cell 0 comma 25 space mol end cell equals cell fraction numerator m subscript Ca C O subscript 3 end subscript over denominator 100 space bevelled gram over mol end fraction end cell row cell m subscript Ca C O subscript 3 end subscript end cell equals cell 0 comma 25 space mol cross times 100 space begin inline style bevelled gram over mol end style end cell row blank equals cell 25 space gram end cell end table 


Dengan demikian, massa 0,25 mol kalsium karbonat Ca C O subscript bold 3 (Ar C= 12, 0=16, Ca = 40) adalah 25 gram.

0

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