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Gas berikut yang mempunyai jumlah molekul sama dengan 5,6 gram gas CO  adalah ....

Pertanyaan

Gas berikut yang mempunyai jumlah molekul sama dengan 5,6 gram gas CO left parenthesis Mr equals space 28 space g space mol to the power of negative sign 1 end exponent right parenthesis space adalah ....

  1. 1 gram gas H subscript 2  left parenthesis Mr equals space 2 space g space mol to the power of negative sign 1 end exponent right parenthesis

  2. 3,2 gram gas O subscript 2  left parenthesis Mr equals 32 space g space mol to the power of negative sign 1 end exponent right parenthesis

  3. 7,1 gram gas  Cl subscript 2 left parenthesis Mr space equals space 71 space g space mol to the power of negative sign 1 end exponent right parenthesis

  4. 12,8 gram gas S O subscript 2  left parenthesis Mr equals space 64 space g space mol to the power of negative sign 1 end exponent right parenthesis

  5. 13,8 gram gas N O subscript 2  left parenthesis Mr space equals space 46 space g space mol to the power of negative sign 1 end exponent right parenthesis

Pembahasan Soal:

X double bond n space x space L  X double bond jumlah space partikel n equals space mol space zat L double bond bilangan space Avogadro

table attributes columnalign right center left columnspacing 0px end attributes row cell n space C O end cell equals cell fraction numerator massa space C O over denominator Mr space C O end fraction end cell row blank equals cell fraction numerator 5 comma 6 space g over denominator 28 space g forward slash mol end fraction end cell row blank equals cell 0 comma 2 space mol end cell row blank blank blank row X equals cell 0 comma 2 space mol space x space 6 comma 02 space x space 10 to the power of 23 end cell row blank equals cell 1 comma 204 space x space 10 to the power of 23 end cell end table

Jumlah partikel CO adalah 1 comma 204 space x 10 to the power of 23.

Berdasarkan hal tersebut di bawah ini adalah penentuan jumlah partikel senyawa pada:

A. 1 gram gas H subscript 2  left parenthesis Mr equals space 2 space g space mol to the power of negative sign 1 end exponent right parenthesis

table attributes columnalign right center left columnspacing 0px end attributes row cell n space H subscript 2 end cell equals cell fraction numerator space 1 space g over denominator 2 space g forward slash mol end fraction end cell row blank equals cell 0 comma 5 space mol space end cell row X equals cell 0 comma 5 space mol space x space 6 comma 02 space x space 10 to the power of 23 end cell row blank equals cell 3 comma 01 space x space 10 to the power of 23 end cell end table 

 

B. 3,2 gram gas O subscript 2  left parenthesis Mr equals 32 space g space mol to the power of negative sign 1 end exponent right parenthesis

table attributes columnalign right center left columnspacing 0px end attributes row cell n space O subscript 2 end cell equals cell fraction numerator 3 comma 2 space g over denominator 32 space g forward slash mol end fraction end cell row blank equals cell 0 comma 1 space mol end cell row X equals cell 0 comma 1 space mol space x space 6 comma 02 space x 10 to the power of 23 end cell row blank equals cell 0 comma 602 space x space 10 to the power of 23 end cell row blank equals cell 6 comma 02 space x space 10 to the power of 22 end cell end table

 

C. 7,1 gram gas  Cl subscript 2 left parenthesis Mr space equals space 71 space g space mol to the power of negative sign 1 end exponent right parenthesis

table attributes columnalign right center left columnspacing 0px end attributes row cell n space Cl subscript 2 end cell equals cell fraction numerator 7 comma 1 space g over denominator 71 space g forward slash mol end fraction end cell row blank equals cell 0 comma 1 space mol end cell row X equals cell 0 comma 1 space mol space x space 6 comma 02 space x 10 to the power of 23 end cell row blank equals cell 0 comma 602 space x space 10 to the power of 23 end cell row blank equals cell 6 comma 02 space x space 10 to the power of 22 end cell end table 

 

D. 12,8 gram gas S O subscript 2  left parenthesis Mr equals space 64 space g space mol to the power of negative sign 1 end exponent right parenthesis

table attributes columnalign right center left columnspacing 0px end attributes row cell n space S O subscript 2 end cell equals cell fraction numerator 12 comma 8 space g over denominator 64 space g forward slash mol end fraction end cell row blank equals cell 0 comma 2 mol end cell row X equals cell 0 comma 2 space mol space x space 6 comma 02 space x 10 to the power of 23 end cell row blank equals cell 1 comma 204 space x space 10 to the power of 23 end cell end table

 

E. 13,8 gram gas N O subscript 2  left parenthesis Mr space equals space 46 space g space mol to the power of negative sign 1 end exponent right parenthesis

table attributes columnalign right center left columnspacing 0px end attributes row cell n space S O subscript 2 end cell equals cell fraction numerator 13 comma 8 space g over denominator 46 space g forward slash mol end fraction end cell row blank equals cell 0 comma 2 mol end cell row X equals cell 0 comma 3 mol space x space 6 comma 02 space x 10 to the power of 23 end cell row blank equals cell 1 comma 806 space x space 10 to the power of 23 end cell end table 

 

Berdasarkan hasil perhitungan di atas, gas S O subscript 2 ternyata mempunyai jumlah partikel yang sama dengan gas CO yaitu sebesar1 comma 204 space x 10 to the power of 23 . 

Jadi, jawaban yang tepat adalah D.

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 04 Mei 2021

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