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Zat berikut yang memiliki jumlah mol terbesar adalah ....(Ar Fe=56, R=0,082 L⋅atm⋅mol−1⋅K−1, dan L=6,02×1023)

Pertanyaan

Zat berikut yang memiliki jumlah mol terbesar adalah ....spacebegin mathsize 14px style left parenthesis Ar space Fe equals 56 comma space R equals 0 comma 082 space L middle dot atm middle dot mol to the power of negative sign 1 end exponent middle dot K to the power of negative sign 1 end exponent comma space dan space L equals 6 comma 02 cross times 10 to the power of 23 right parenthesis end style

  1. 3 gram asam asetatspacebegin mathsize 14px style open parentheses C H subscript 3 C O O H close parentheses end style

  2. begin mathsize 14px style 3 comma 01 cross times 10 to the power of 22 end style atom besispace

  3. 5,6 liter gas CO keadaan standarspace

  4. 2,46 liter gasspacebegin mathsize 14px style N H subscript 3 space left parenthesis 27 degree C comma space 2 space atm right parenthesis end style

  5. 5 liter gas begin mathsize 14px style O subscript 2 end style dan pada suhu dan tekanan yang sama, 1 liter gas begin mathsize 14px style H subscript 2 end style massanya 4 gramspace

I. Solichah

Master Teacher

Jawaban terverifikasi

Jawaban

jawaban yang benar adalah E.space

Pembahasan

Untuk dapat menyelesaikan soal di atas, maka perlu dicari satu persatu jumlah mol dalam tiap senyawa.

a. 3 gram asam asetat


table attributes columnalign right center left columnspacing 0px end attributes row cell mol space C H subscript 3 C O O H end cell equals cell fraction numerator massa over denominator Mr space C H subscript 3 C O O H end fraction end cell row blank equals cell fraction numerator massa over denominator 2 cross times A subscript r space C plus 2 cross times A subscript r space O plus 4 cross times A subscript r space H end fraction end cell row blank equals cell fraction numerator 3 space g over denominator left parenthesis 2 cross times 12 plus 2 cross times 16 plus 4 cross times 4 right parenthesis space g forward slash mol end fraction end cell row blank equals cell equals fraction numerator 3 space g over denominator 60 space g forward slash mol end fraction end cell row blank equals cell 0 comma 05 space mol end cell end table 
 

b. 3 comma 01 cross times 10 to the power of 22 atom besi


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 mol space Fe end cell equals cell fraction numerator Jumlah space partikel over denominator L end fraction end cell row blank equals cell fraction numerator 3 comma 01 cross times 10 to the power of 22 over denominator 6 comma 02 cross times 10 to the power of 23 end fraction end cell row blank equals cell 0 comma 05 space mol end cell end table 
 

c. 5,6 liter gas CO (STP)


table attributes columnalign right center left columnspacing 0px end attributes row cell V subscript STP end cell equals cell mol cross times 22 comma 4 space L end cell row mol equals cell fraction numerator V subscript STP over denominator 22 comma 4 end fraction equals fraction numerator 5 comma 6 space L over denominator 22 comma 4 space L end fraction end cell row blank equals cell 0 comma 25 space mol end cell end table 
 

d. 2,46 liter gas N H subscript 3 space left parenthesis 27 degree C comma space 2 space atm right parenthesis


table attributes columnalign right center left columnspacing 0px end attributes row cell P cross times V end cell equals cell mol cross times R cross times T end cell row mol equals cell fraction numerator P cross times V over denominator R cross times T end fraction end cell row blank equals cell fraction numerator 2 atm cross times 2 comma 46 space L over denominator 0 comma 082 space L point atm point mol to the power of negative sign 1 end exponent. K to the power of negative sign 1 end exponent cross times 300 space K end fraction end cell row blank equals cell fraction numerator 4 comma 92 over denominator 24 comma 6 end fraction equals 0 comma 2 space mol end cell end table 
 

e. 5 liter gas O subscript 2 pada keadaan yang sama dengan 1 liter H subscript 2 4 gram


table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator mol space O subscript 2 over denominator V space O subscript 2 end fraction end cell equals cell fraction numerator mol space H subscript 2 over denominator V space H subscript 2 end fraction end cell row cell fraction numerator mol space O subscript 2 over denominator 5 space L end fraction end cell equals cell fraction numerator begin display style fraction numerator massa over denominator Mr space H subscript 2 end fraction end style over denominator 1 space L end fraction end cell row cell fraction numerator mol space O subscript 2 over denominator 5 space L end fraction end cell equals cell fraction numerator begin display style fraction numerator 4 space gram over denominator 2 space gram forward slash mol end fraction end style over denominator 1 space L end fraction end cell row cell mol space O subscript 2 end cell equals cell 5 over 1 cross times 2 space mol end cell row blank equals cell 10 space mol end cell end table 
 

Dengan demikian maka mol terbesar adalah mol gas oksigen.

Oleh karena itu, jawaban yang benar adalah E.space

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