tugas analisis farmasi ii (analisis sediaan padat)

7/25/2019 Tugas Analisis Farmasi II (Analisis Sediaan Padat)

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TUGAS ANALISIS FARMASI II

ANALISIS BAHAN AKTIF DALAM SEDIAAN FARMASI BENTUK PADAT

NAMA : YANISH ADITYA MAHENDRA

NIM : 051311133236

KELAS : C1.

Jawaban:

AS-SAL, BM=138, A 1% = 552

E= A1 x BM 10

= 552 x 13810

= 7617,6 grl / L

A= A1

10000= 552

10000= 0 , 0552 mg / L

AMDKAIN, BM=460, E=1150

A1 = E x 10

BM = 1150 x 10

460= 25 g /100 ml

A= E

1000 BM = 1150

1000 x 460= 0,0025 mg / L

H !, BM=80, A=0,050 A

1 = A x10000 = 0,050 x 10000 = 500 g /100 ml

E= A x1000 BM = 0,050 x1000 x 80 = 4000 grl / L

DHS, BM=60, A 1% = 12

E= A1 x BM

10= 12 x 60

10= 72 grl / L

A= A1

10000 = 12

10000 = 0,0012 mg / L

Ana"#$BM

A1%1&'100

(")

E M)1&*"'

")

A ++()1(&'"

AS.SAL13

8552

AMDKAIN

460

1150

H ! 80 0,050

DHS 60 12


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NIM : 051311133236

KELAS : C

2. 60 (& $ab" $ A"+* n " " a/ 100,0 (" /#+#+ $ 10,0 (" a/ 50,0 ("

/#+ * " ab *ban A = 0,0025. Ka/a* A"+ n " " /a"a( $ab" $ b a* 50 (& Jawaban:A1% = 65 g / 100 ml a = 0,0065 (&'L

C = A

a = 0,0025

0,0065= 0,3846 mg

L /a"a( 50,0 (")

C =5010 x0,3846 = 1,9230

mg L /a"a( 10,0 (")

C = 1,9230mg L

= 1,9230 mg1000 ml

= 0,1923 mg100 ml

%recovery =0,1923 mg

50 mg x100 = 0,3846

ZATKADAR(ppm

A p!"!

#m!$E

A p!"!

#m!$N

E $*a/# " E) 2 0,200 0

N & $* "N) 2 0 0,400

Sa(+ "

a(+ *anE N 0,180 0,360

3. Jawaban: K (+ n n E =

ae = AsteCste

= AspeCspe

Cspe = Aspe.Cste

Aste = 0,180.2

0,200= 1,8 ppm

K (+ n n N =

an = AstnCstn

= AspnCspn


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NIM : 051311133236

KELAS : CCspn = Aspn . Cstn

Astn = 0,360.2

0,400= 1,8 ppm

%!"& '!"! )*+ !"&,- "!-!m .!mp/ ! *) )*! 1 ppm "! '!"!,*4)*+ )- "!-!m .!mp/ ! *) )*! 1 ppm

4. Jawaban:

A = a . a .

0,800 = 0,150 . 0,060 . A = a . a .

0,500 = 0,050. 0,200 .

0,800 = 0,150. 0,060. 10,500 = 0,050. 0,200. 3

0,800 = 0,150. 0,060.1,500 = 0,150. 0,600. 7 -0, 00 = -0,540.

= 1,2963 ppm

0,800 = 0,150. 0,06.1,2963)0,800 = 0,150. 0,0 8

Cx = (0,800 0,0778 )0,150

= 0,72220,150

= 4,8147 ppm

%!"& '!"! p! !*)+!m,- "!-!m .!mp/ ! *) )*! 1 ppm "!'!"! ',7)& "!-!m .!mp/ ! *) )*! 1 2863 ppm


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NIM : 051311133236

KELAS : C

5. Jawaban:

Qy= ayyayiso

= 0,47310

x 10,461

= 0,1026

Qx= axxaxiso

= 0,52810

x 10,461

= 0,1145

Qox= Acx Aciso

= 0,7070,469

= 1,5075

Qoy = Acy Aciso

= 0,5480,469

= 1,1684

Cx=(Qox Qy )(Qx Qy )

Acisoaxx

= (1,5075 0,1026 )(0,1145 0,1026 ) 0,4690,0528

= 1048,66 ppm

Cy=(Qoy Qx )(Qx Qy )

Acisoayy

= (1,1684 0,1145 )(0,1145 0,1026 ) 0,4690,0473

= 878,14 ppm

%!"& '!"! AS "!-!m .!mp/ ! *) )*! 10 66 ppm "! '!"!AB "!-!m .!mp/ ! *) )*! 1 ppm


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NIM : 051311133236

KELAS : C

6. Jawaban: K (+ n n +a*a $a( " =

A+ = ;0,645-0,566; = 0,0 9

a Ap= 0,07910

= 0,0079

A + = ;0, 84-0, 03; = 0,081

Cp= ACpa Ap

= 0,0810,0079

= 10,25 ppm

K (+ n n < #n = A< = ;0,040-0,052; = 0,012

a Ak = 0,0121

= 0,012

A < = ;0,836-0,843; = 0,00

Ck = ACk a Ak

= 0,0070,012

= 0,58 ppm

%!"& '!"! p! !*)+!m,- "!-!m .!mp/ ! *) )*! 10 25 ppm "!'!"! ',7)& "!-!m .!mp/ ! *) )*! 0 5 ppm


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NIM : 051311133236

KELAS : C

.

Jawaban:

Ay= A maksy mAy1 +nAy2m+n

= 0,656 (5 0,752 )+(5 0,549 )5 +5 =

0,0055

a Ay = AyCy

= 0,00551

= 0,0055

ACy= 0,654 (5 0,749 )+ (5 0,545 )

5 + 5= 0,0070

Cy= ACya Ay

= 0,00700,0055

= 1,2727 ppm

%!"& '!"! N&p!*,- "!-!m .!mp/ ! *) )*! 1 2 2 ppm

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