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Tentukan massa atom relatif (Ar) dari 3,25 gram logam X yang mempunyai jumlah partikel  atom X!

Pertanyaan

Tentukan massa atom relatif (Ar) dari 3,25 gram logam X yang mempunyai jumlah partikel 3 comma 01 cross times 10 to the power of 22 atom X!
 

Pembahasan Soal:

N double bond n cross times 6 comma 02 cross times 10 to the power of 23 3 comma 01 cross times 10 to the power of 22 equals equals n cross times 6 comma 02 cross times 10 to the power of 23 n equals 0 comma 05 space mol  Ar space X equals bevelled fraction numerator 3 comma 25 over denominator 0 comma 05 end fraction Ar space X equals 65 space begin inline style bevelled g over mol end style

Dengan demikian, maka jawaban yang tepat adalah 65 g/mol.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

A. Budi

Mahasiswa/Alumni Universitas Pendidikan Indonesia

Terakhir diupdate 30 April 2021

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Pertanyaan yang serupa

Unsur boron mempunyai dua isotop di alam, yaitu B-10 dengan kelimpahan 20 % dan B-11 dengan kelimpahan 80 %. Jika massa atom B-10 sebesar 10 sma dan massa atom B-11 sebesar 11 sma, massa atom relatif ...

Pembahasan Soal:

massa B-10 = 10

massa B-11 = 11

table attributes columnalign right center left columnspacing 0px end attributes row cell Ar space B end cell equals cell left parenthesis kelimpahan space B minus sign 10 space x space massa space B minus sign 10 right parenthesis space plus space left parenthesis kelimpahan space B minus sign 11 space x space massa space B minus sign 11 right parenthesis space end cell row blank equals cell left parenthesis 20 space percent sign space x space 10 right parenthesis plus space left parenthesis 80 space percent sign space x space 11 right parenthesis end cell row blank equals cell 2 plus 8 comma 8 end cell row blank equals cell 10 comma 8 end cell end table

Sehingga massa atom relatif boron adalah 10,8.

Jadi, jawaban yang tepat adalah C.

4

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

​​​Jika massa 1 atom C-12 =  kg dan massa 1 atom Na =  kg, massa atom relatif Na adalah ....

Pembahasan Soal:

Untuk mencari massa atom relatif Na, dapat digunakan persamaan berikut.
 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell A subscript r space Na end cell equals cell space fraction numerator massa space rata bond rata space 1 space atom space Na over denominator begin display style 1 over 12 space x space massa space 1 space atom space C minus sign 12 end style end fraction end cell row cell A subscript r space Na end cell equals cell fraction numerator 3 comma 81 space x space 10 to the power of negative sign 26 end exponent over denominator begin display style 1 over 12 end style x space 1 comma 99 space x space 10 to the power of negative sign 26 end exponent space end fraction end cell row cell A subscript r space Na end cell equals cell 22 comma 97 end cell row cell A subscript r space Na end cell equals 23 end table end style  
 

Jadi, jawaban yang benar adalah D.undefined 

 

0

Roboguru

Diketahui massa atom relatif (Ar) Na = 23. Tentukan massa rata-rata dari 1 atom natrium jika dinyatakan dalam sma ...

Pembahasan Soal:

Massa atom relatif adalah perbandingan massa antara atom yang satu terhadap atom yang lainnya. Massa pembanding yang telah disepakati adalah 1 over 12 dari massa 1 atom C-12. Oleh karena umumnya unsur terdiri atas beberapa isotop, maka pada penetapan massa atom relatif digunakan massa rata-rata dari isotop-isotopnya.
 

table attributes columnalign right center left columnspacing 0px end attributes row cell Ar space unsur space Na end cell equals cell fraction numerator massa space rata bond rata space 1 space atom space unsur space Na over denominator 1 space sma end fraction end cell row 23 equals cell fraction numerator massa space rata bond rata over denominator 1 space sma end fraction end cell row cell massa space rata bond rata end cell equals cell 23 cross times 1 end cell row blank equals cell 23 space sma end cell end table 
 

Jadi, dapat disimpulkan jawaban yang tepat yaitu 23 sma.space

0

Roboguru

Hitunglah massa hidrogen dalam 60 gram urea !

Pembahasan Soal:

begin mathsize 14px style m space H equals fraction numerator 4 cross times italic A subscript r italic space H over denominator italic M subscript r space C O open parentheses N H subscript 2 close parentheses subscript 2 end fraction cross times m space C O open parentheses N H subscript 2 close parentheses subscript 2 m space H equals fraction numerator 4 cross times 1 over denominator 60 end fraction cross times 60 space gram m space H equals 4 space gram end style  
 

Jadi, massa hidrogen adalah 4 gram.space  

0

Roboguru

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