Fill In the table and answer the following question. What is the n-propanol conc

Question

Fill In the table and answer the following question.

What is the n-propanol concentration in each of the above solution? Density of n-propanol is 0.8 g/mL.

Final (EtOH] 2. References and standard solutions in 25 ml volumetric flasks Flask # First, add.... Then dilute to 25 ml with 1 | 2.5 mL n-prop/H;0 DI water 2 | 2.5 ml n-prop/EtOH/H;0 the ethanol/H20 stock solution Flask # First, add.... Then, add ? mL 10 mg/mL EtOH/H;0 3 2.5 ml n-prop/H;0 Dilute with DI | 4 || 2.5 mL n-prop/H,0 water to 100 mL 5 2.5 mL n-prop/H;0 6 || 2.5 mL n-prop/H;0 7 | 2.5 mL n-prop/H,0 o.a. 0.2 mg/mL 0.5 mg/mL 0.8 mg/mL 1.0 mg/mL 1.5 mg/mL

Explanation / Answer

density of propanol =0.8 menas , 1 mL of propanol weight is 0.8 g

1 > 2.5 mL of propanol = 0.8 * 2.5 g= 2.00 g of propanol

now 25 mL of solution contains 2 g of propanol

1 mL of solution contains 2/25 = 0.08 g/mL propanol

so propanol concentraiton is 0.08 g/mL

2 > same as above

3> for caluclation of ? mL of 10 mg/mL of EtOH,

let x mL of 10 mg / mL of EtOH is mixed to form 0.2 mg/mL of final solution EtOH concentration.

now , 10 mg / mL menas 1 mL contains 10 mg

so x ml contais 10x mg of EtOH.

now 0.2 mg/mL menas 1 mL of slution contains 0.2 mg of ethanol

now total volume is 100+2.5 = 102.5 mL of solution contains 0.2 * 102.5 = 20.5 mg of EtOH final solution

now 10x = 20.5

x = 2.05

so 2.05 mL of 10 mg/mL is needed

now density of n propanol = 0.8 g/mL

2.5 mL of propanol = 2.5 * 0.8 = 2.0 g of propanol

now (100+2.05 ) mL contains 2 g of propanol

1 mL contains 2/ 102.05 = 0.0195 g/mL = 19.5 mg/mL

answer 4> similarly as from the above process

10x = 102.5 * 0.5

x = 5.12

so 5.12 ml of 10 mg/mL EtOH is needed

now n- propanol concentration is equal to 2/105.12 = 0.01902 g/mL = 19.02 mg/mL

answer 5

similarly 10x = 102.5 * 0.8

x= 8.2

so 8.2 mL of 10 mg/mL of EtOH is needed

now concentration of n propanol = 2/108.2 = 0.01848 g / mL = 18.48 mg/mL

answer 6 >

similarly 10 x = 102.5 * 1.0

x=10.25

so 10.25 mL of 10 mg/ mL EtOH is required

concentration of n- propanol = 2/ 110.25 = 18.14 mg/mL

answer 7

10x = 102.5* 1.5

x= 15.37

15.37 mL 10mg/mL EtOH required

concentration of n propanol = 2 / 115.37 = 17.33 mg / mL

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