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Dalam suatu percobaan terdapat larutan garam NaCl 10% (massa/massa), jika Mr NaCl = 58,5 dan Mr H subscript 2 O = 18 dan massa jenis air = massa jenis garam = 1 gram / ml. Tentukanlah:

A. Konsentrasi Molaritas larutan garam tersebut

B. Konsentrasi Molalitas larutan garam tersebut

C. Fraksi mol garam dalam larutan garam tersebut

D. Fraksi mol Air dalam larutan garam tersebut

G. Suprobo

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jawaban sesuai dengan uraian di atas.

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table attributes columnalign right center left columnspacing 0px end attributes row cell massa space Na Cl end cell equals cell 10 percent sign left parenthesis massa forward slash massa right parenthesis end cell row cell Mr space Na Cl end cell equals cell 58 comma 5 space g forward slash mol end cell row cell Mr space H subscript 2 O end cell equals cell 18 space g forward slash mol end cell row cell rho space air end cell equals cell rho space garam equals 1 space g forward slash ml end cell end table

NaCl 10% artinya dalam 100 gram larutan NaCl terdapat NaCl murni 10% x 100 gram= 10 gram NaCl. Massa air dalam 100 gram larutan adalah (100 gram-10 gram = 90 gram air).

A. Konsentrasi molaritas larutan garam 

table attributes columnalign right center left columnspacing 0px end attributes row cell Molaritas space Na Cl end cell equals cell fraction numerator rho space x space 10 space x space percent sign over denominator Mr end fraction end cell row blank equals cell fraction numerator 1 space x space 10 space x space 10 over denominator 58 comma 5 end fraction end cell row blank equals cell 1 comma 709 space M end cell row blank blank blank end table

B. Konsentrasi molalitas larutan garam 

table attributes columnalign right center left columnspacing 0px end attributes row cell molalitas space Na Cl end cell equals cell massa over Mr space x space fraction numerator 1000 over denominator massa space pelarut end fraction end cell row blank equals cell fraction numerator 10 space g over denominator 58 comma 5 space g forward slash mol end fraction x space 1000 over 90 end cell row blank equals cell 1 comma 89 space m end cell end table

C. Fraksi mol garam dalam larutan garam 

table attributes columnalign right center left columnspacing 0px end attributes row nt equals cell fraction numerator 10 space g over denominator 58 comma 5 space g forward slash mol end fraction equals 0 comma 17 space mol end cell row np equals cell fraction numerator 90 space g over denominator 18 space g forward slash mol end fraction equals 5 space mol end cell row Xt equals cell fraction numerator nt over denominator nt and np end fraction end cell row blank equals cell fraction numerator 0 comma 17 over denominator 0 comma 17 plus 5 end fraction end cell row blank equals cell 0 comma 032 end cell end table

D. Fraksi mol air dalam larutan garam

table attributes columnalign right center left columnspacing 0px end attributes row nt equals cell fraction numerator 10 space g over denominator 58 comma 5 space g forward slash mol end fraction equals 0 comma 17 space mol end cell row np equals cell fraction numerator 90 space g over denominator 18 space g forward slash mol end fraction equals 5 space mol end cell row Xp equals cell fraction numerator np over denominator np and nt end fraction end cell row blank equals cell fraction numerator 5 over denominator 5 plus 0 comma 17 end fraction end cell row blank equals cell 0 comma 967 space end cell end table

Jadi, jawaban sesuai dengan uraian di atas.

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