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Suatu garam  mempunyai. Jika dalam 250 mL larutan jenuhnya mengandung 0,5825 mg , massa atom relatif L adalah ....

Pertanyaan

Suatu garam LSO subscript 4 mempunyaispace Ksp equals 1 comma 0 cross times 10 to the power of negative sign 10 end exponent. Jika dalam 250 mL larutan jenuhnya mengandung 0,5825 mg LSO subscript 4, massa atom relatif L adalah ....

  1. 233 space g space mol to the power of negative sign 1 end exponent 

  2. 207 space g space mol to the power of negative sign 1 end exponent 

  3. 137 space g space mol to the power of negative sign 1 end exponent 

  4. 88 space g space mol to the power of negative sign 1 end exponent 

  5. 40 space g space mol to the power of negative sign 1 end exponent 

Pembahasan Soal:

Untuk menghitung massa atom relatif unsur L, dapat dilakukan dengan langkah-langkah berikut ini:

Langkah pertama, menghitung kelarutan (s) dari LSO subscript 4:


table attributes columnalign right center left columnspacing 0px end attributes row cell LSO subscript 4 end cell rightwards harpoon over leftwards harpoon cell L to the power of 2 plus sign plus S O subscript 4 to the power of 2 minus sign end exponent end cell row blank blank cell space space s space space space space space space space space space space s space space space space space space space space s end cell row cell K subscript sp end cell equals cell 1 comma 0 cross times 10 to the power of negative sign 10 end exponent end cell row cell K subscript sp end cell equals cell open square brackets L to the power of 2 plus sign close square brackets open square brackets S O subscript 4 to the power of 2 minus sign end exponent close square brackets end cell row blank equals cell open square brackets s close square brackets open square brackets s close square brackets end cell row blank equals cell s squared end cell row cell 1 comma 0 cross times 10 to the power of negative sign 10 end exponent end cell equals cell s squared end cell row s equals cell square root of 1 comma 0 cross times 10 to the power of negative sign 10 end exponent end root end cell row blank equals cell 10 to the power of negative sign 5 end exponent end cell end table  


Langkah kedua, menghitung mol LSO subscript 4:


table attributes columnalign right center left columnspacing 0px end attributes row cell M space LSO subscript 4 end cell equals cell s space LSO subscript 4 end cell row blank equals cell 10 to the power of negative sign 5 end exponent end cell row V equals cell 250 space mL equals 0 comma 25 space L end cell row M equals cell n over V end cell row n equals cell M cross times V end cell row blank equals cell 10 to the power of negative sign 5 end exponent space M cross times 0 comma 25 space L end cell row blank equals cell 2 comma 5 cross times 10 to the power of negative sign 6 end exponent space mol end cell end table


langkah ketiga, menghitung M subscript r space LSO subscript 4:


table attributes columnalign right center left columnspacing 0px end attributes row massa equals cell 0 comma 5825 space mg end cell row blank equals cell 0 comma 5825 cross times 10 to the power of negative sign 3 end exponent space g end cell row n equals cell massa over M subscript r end cell row cell 2 comma 5 cross times 10 to the power of negative sign 6 end exponent space mol end cell equals cell fraction numerator 0 comma 5825 cross times 10 to the power of negative sign 3 end exponent space g over denominator M subscript r end fraction end cell row cell M subscript r end cell equals cell fraction numerator 0 comma 5825 cross times 10 to the power of negative sign 3 end exponent space g over denominator 2 comma 5 cross times 10 to the power of negative sign 6 end exponent space mol end fraction end cell row blank equals cell 233 space g space mol to the power of negative sign 1 end exponent end cell end table  


Langkah keempat, menghitung A subscript r space L:


table attributes columnalign right center left columnspacing 0px end attributes row cell A subscript r space L plus space A subscript r space S plus 4 left parenthesis A subscript r space O right parenthesis end cell equals cell 233 space g space mol to the power of negative sign 1 end exponent end cell row cell A subscript r space L plus 32 plus 4 left parenthesis 16 right parenthesis end cell equals cell 233 space g space mol to the power of negative sign 1 end exponent end cell row cell A subscript r space L end cell equals cell left parenthesis 233 minus sign 32 minus sign 64 right parenthesis space g space mol to the power of negative sign 1 end exponent end cell row blank equals cell 137 space g space mol to the power of negative sign 1 end exponent end cell end table


Oleh karena itu, massa atom relatif L adalah 137 space g space mol to the power of negative sign 1 end exponent.

Jadi, jawaban yang benar adalah C.

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 19 Mei 2021

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