Question

What is the concentration of EACH solution in these reactions...

Table 1: Concentrations Used of H2O2 and KI
Trial	H2O2	KI
1	5 mL of 3% H2O2	10 mL of 0.60M KI
2	5 mL of 3% H2O2	10 mL of 0.30M KI
3	5 mL of 2.25% H2O2	10 mL of 0.60 M KI

Answer 1

1. 5 mL OF 3% H₂O₂= First we will calculate the actual grams of H2O2 Presene in 5 mL solution,

so, 5*3=15/100=.15 grams of H2O2 PRESENT IN 5 mL

MOL. WT of H2O2 =34 therefore we willl calculate moles of H2O2=0.15/34=0.0044 MOLES.

10 mL .60M KI= MOL. WT OF KI=166,

0.60*166=99.6 FOR 1000 ML, THEN,

FOR 100 ML=9.96 GMS,THEN FOR 10ML=?

10=0.996 GMS KI present in 10 mL SOLUTION,

THEN % OF KI = WT BY VOLUME=0.996/10=9.96% AND WT /WT=0.996/9.04=11%

2. 5 mL OF 3% H₂O₂= First we will calculate the actual grams of H2O2 Presene in 5 mL solution,

so, 5*3=15/100=.15 grams of H2O2 prenent in 5 mL.

MOL. WT of H2O2 =34 therefore we willl calculate moles of H2O2=0.15/34=0.0044 MOLES.

10 mL 0.30 M KI SOLUTION = MOL WT OF KI=166,

0.30 *166=49.8 FOR 1000 ML THEN, FOR 100 ML 4.98 GMS IS REQUIRED,

THEN FOR 10 ML 0.498 GMS OF KI IS REQUIRED.

% OF KI PRESENT= , WT/WT=04.98/9.502=5.24 % KI, WT/VOLUME=0.498/10=4.98% KI.

3. 5 mL of 2.25 % H2O2= FIRST WE WILL CALCULATE GMS OF H2O2 USING FORMULA=

5*2.25=11.25/100=.0.1125 GMS,MOLES OF H2O2 = 0.1125/34.01=0.0033 MOLES OF H2O2.

10 mL .60M KI= MOL. WT OF KI=166,

0.60*166=99.6 FOR 1000 ML, THEN,

FOR 100 ML=9.96 GMS,THEN FOR 10ML=?

10=0.996 GMS KI present in 10 mL SOLUTION,

THEN % OF KI = WT BY VOLUME=0.996/10=9.96% AND WT /WT=0.996/9.04=11%