suatu larutan di buat dengan cara melarutkan 15,2 gram NaCl dalam 197 gram air . jika larutan itu

Pertanyaan:

suatu larutan di buat dengan cara melarutkan 15,2 gram NaCl dalam 197 gram air . jika larutan itu mempunyai massa jenis 1,012 gram/mol tentukan fraksi mol NaCl

Massa molar (Mr) NaCl adalah 58,44 g/mol. Jumlah mol NaCl ( n N a C l n sub cap N a cap C l end-sub n N a C l ) dihitung menggunakan rumus: n N a C l = massa NaCl Mr NaCl n sub cap N a cap C l end-sub equals the fraction with numerator massa NaCl and denominator Mr NaCl end-fraction n N a C l = m a s s a N a C l M r N a C l n N a C l = 15 , 2 g 58 , 44 g/mol n sub cap N a cap C l end-sub equals the fraction with numerator 15 comma 2 g and denominator 58 comma 44 g/mol end-fraction n N a C l = 1 5 , 2 g 5 8 , 4 4 g / m o l n N a C l ≈ 0 , 26 mol n sub cap N a cap C l end-sub is approximately equal to 0 comma 26 mol n N a C l ≈ 0 , 2 6 m o l

Massa molar (Mr) air ( H 2 O cap H sub 2 cap O H 2 O ) adalah 18,02 g/mol. Jumlah mol air ( n H 2 O n sub cap H sub 2 cap O end-sub n H 2 O ) dihitung menggunakan rumus: n H 2 O = massa air Mr air n sub cap H sub 2 cap O end-sub equals the fraction with numerator massa air and denominator Mr air end-fraction n H 2 O = m a s s a a i r M r a i r n H 2 O = 197 g 18 , 02 g/mol n sub cap H sub 2 cap O end-sub equals the fraction with numerator 197 g and denominator 18 comma 02 g/mol end-fraction n H 2 O = 1 9 7 g 1 8 , 0 2 g / m o l n H 2 O ≈ 10 , 93 mol n sub cap H sub 2 cap O end-sub is approximately equal to 10 comma 93 mol n H 2 O ≈ 1 0 , 9 3 m o l

Fraksi mol NaCl ( X N a C l cap X sub cap N a cap C l end-sub X N a C l ) adalah perbandingan jumlah mol NaCl dengan jumlah mol total dalam larutan (mol NaCl + mol air). Mol total = n N a C l + n H 2 O Mol total equals n sub cap N a cap C l end-sub plus n sub cap H sub 2 cap O end-sub M o l t o t a l = n N a C l + n H 2 O Mol total ≈ 0 , 26 mol + 10 , 93 mol Mol total is approximately equal to 0 comma 26 mol plus 10 comma 93 mol M o l t o t a l ≈ 0 , 2 6 m o l + 1 0 , 9 3 m o l Mol total ≈ 11 , 19 mol Mol total is approximately equal to 11 comma 19 mol M o l t o t a l ≈ 1 1 , 1 9 m o l Rumus fraksi mol NaCl: X N a C l = n N a C l Mol total cap X sub cap N a cap C l end-sub equals the fraction with numerator n sub cap N a cap C l end-sub and denominator Mol total end-fraction X N a C l = n N a C l M o l t o t a l X N a C l = 0 , 26 mol 11 , 19 mol cap X sub cap N a cap C l end-sub equals the fraction with numerator 0 comma 26 mol and denominator 11 comma 19 mol end-fraction X N a C l = 0 , 2 6 m o l 1 1 , 1 9 m o l X N a C l ≈ 0 , 023 cap X sub cap N a cap C l end-sub is approximately equal to 0 comma 023 X N a C l ≈ 0 , 0 2 3

Fraksi mol NaCl dalam larutan tersebut adalah 0,023 . Data massa jenis yang diberikan tidak diperlukan untuk perhitungan fraksi mol.