3.6 x 10 A 9.0 x 1 B, 224 10\' 8 x 10 c. (3x 10-36 x 10)- D. (2.4x 105(5.6 x 103

Question

3.6 x 10 A 9.0 x 1 B, 224 10' 8 x 10 c. (3x 10-36 x 10)- D. (2.4x 105(5.6 x 103)- 2. Solve the following dilutions. Express as both a ratio (e.g. 1:10) and in exponential form (e g 101). Diluate Diluent 27 mls. 99 mls. 49.5 mls. 99.9 mls. 4.5 mls Ratio A. 3 mls. B. 1 ml C. 0.5 ml D. 0.1 ml E. 10.5ml 3. 3 mls. bacterial suspension plus 6 mls. water gives a dilution. 4. 4. How many mls. bacterial suspension would you add to 32 mls. water to get a 1:9 dilution? 5. 5. 1.8 mls. bacterial suspension would be added to mls. broth to obtain a 1:6 dilution. 1 ml 1 ml 1 ml 6. a. The dilution in in tube B is b·The dilution in tube C is nch resu+s 3 cqvations 9 ml 9 ml 9 ml

Explanation / Answer

2) To calculate the dilution we use a simple formula , Dilution= diluate / (volume of diluate and volume of diluent)

Diluate means that liquid which is to be diluated for example if we make a orange juice dilution by mixing orange juice with water , so ,in this case orange juice is diluate and water is diluent.

1/(1+99)=1/100

3) 3ml of bacterial suspension + 6 ml of water = 9 ml of water

in this case bacterial suspension is diluate and water is diluent so ,by using formula

dilution = 3ml/ 9ml = 1/3 = 0.33333

4) In this question let assume the volume of diluate means bacterial suspension = x ml

volume of water i.e diluent as given to the question is = 32 ml

so, total volume = (x+32)

now dilution is given that is 1:9 ,so using formula we get

Dilution= vol.of diluate/ total volume

or,1/9 = x / x+32

or, x+32= 9x

or, 9x-x = 32

or, 8x = 32

or, x= 4

so, our answer is 4ml that means 4ml of bacterial suspension is added to 32 ml of water to make 1:9 dilution

5) in this quention volume of diluate i.e bacterial suspension is given which is = 1.8 ml

Now, let assume the volume of diluent i.e water = x ml

total volume = (x+ 1.8 ) ml

Dilution is also given which is 1:6 , using the formula it would be

Dilution = 1.8 / (1.8+ x)

or, 1:6 = 1.8 / (1.8+ x) ( putting the dilution value as given to the question)

or, 1/6 = 1.8 / (1.8 + x)

or, x = 9.8

So, for 1:6 dilution of 1.8 ml of bacterial suspension 9.8 ml of water is to addad.

6) This process is called serial dilution as given to the question .

for tube A ,Dilution is = 1ml/ (1+9) ml= 1/10

for tube B dilution is = 1/10 * 1ml/( 1+9) ml

= 1/10 * 1/10

= 1/100

[here ,product is done because for the tube B the diluate is taken from tube A which dilution is 1/10 ,here diluate is not taken from original liquid . for example , one bottle of squash is an aliquote that is original mixture , we now if mix 1ml of orange squash and 9 ml of water forming dilution 0.1 , Now if we take 1 ml of the orange squash mixture of 0.1 dilution with again 9 ml of water then it call serial dilution the formula will be = dilution of the diluate * new dilution ]

For tube C dilution = 1/10 *1/10 * 1/( 1+9)

= 1/1000

(7) Dilution of tube A = 0.5 / ( 0.5 + 49.5) = 1/100

Dilution of tube B = 1/100 * 1ml/ ( 1+9) ml

=1/100 * 1/10

=1/ 1000

Dilution of tube C = 1/1000 * 1ml /(1 +99) ml

= 1/1000 * 1/ 100 = 1/100000

Dilution of tube D = 1/100000 * 1ml / ( 1+ 9 )ml

= 1/100000 * 1/ 10 = 1/1000000

8) Dilution of tube A = 1ml / ( 1+9) ml = 1/10

Dilution of tube B = 1/10 * 1ml/ (1+9) ml

= 1/10 * 1/10 = 1/100

Dilution of tube C = 1/100 * 1ml / ( 1+99) ml = 1/100 * 1/100 = 1/10000

Dilution of tube D = 1/10000 * 1ml / ( 1+9) ml = 10000 * 1/10 = 1/ 100000

Diluate (ml) Diluent(ml) Math Ratio Exponential A 3 27 3/(3+27)=3/30=1/10 1:10 10-1 B 1 99

1/(1+99)=1/100

1:100 10-2 C 0.5 49.5 0.5/(0.5+49.5)=0.01=1/100 1:100 10-2 D 0.1 99.9 0.1/(99.9+0.1)=0.001=1/1000 1:1000 10-3 E 0.5 4.5 0.5/(0.5+4.5)= 0.1=1/10 1:10 10-1

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