Approaches

Remove misconception about making solutions

A solution that is 20% sugar is made sweeter by doubling the amount of sugar. What is the percent of sugar in new solution?
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Be careful: wrong approach followed by most
Original solution: 20 gm sugar in 100 ml d/w
New solution: 40 gm sugar in 100 ml d/w
Percent of sugar in new solution: 40%
incorrect never do like this
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Correct approach
Original solution: 20 gm sugar in 80 ml d/w
New solution: 40 gm sugar in 80 ml d/w
Percent of sugar in new solution:
33.33%

Now, let’s try another
Prepare 1% w/v sodium chloride (anhydrous) solution
1 gm sodium chloride (anhydrous) dissolved in small amount of water and final volume made 100 ml by adding d/w in volumetric flask of 100 ml (Read lower meniscus. Upper meniscus is read for colored solution).

Prepare 1% w/v sodium acetate (anhydrous) solution
1 gm sodium acetate (anhydrous) dissolved in small amount of water and final volume made 100 ml by adding d/w in volumetric flask of 100 ml.

What about preparing 1% w/v sodium acetate tri hydrate
It contains additional 3 molecules of water
Mol. Wt. of sodium acetate: 82
Mol. Wt. of 3 H20: 54
Total molecular wt of sodium acetate tri hydrate is 136
Therefore, 1.65 gm sodium acetate tri hydrate dissolved in small amount of water and final volume made 100 ml in volumetric flask by adding d/w.

Prepare 70% v/v ethyl alcohol from absolute ethyl alcohol
70 ml absolute ethyl alcohol taken in volumetric flask and d/w added to make final volume 100 ml. (about 30 ml d/w added to make final volume 100 ml).

Preparation of 1,000 ml of 10,000 PPM solution of sodium hypochlorite (5% available chlorine)
V1S1 = V2S2
V1 = 1,000 * 10,000 / 50,000 = 200
Therefore, 200 ml sodium hypochlorite mixed with d/w to make final volume 1000 ml in volumetric flask.


Molar solutions
Sodium chloride:
Mol. Wt. = 58.4
58.4 gm dissolved in d/w to make final volume 1 liter.
i.e., 58.4 gm in 1,000 ml = 1 mol/lt
5.84 gm in 100 ml = 1 mol/lt
0.584 gm in 100 ml = 0.1 ml/lt

Potassium hydroxide:
Mol. Wt. = 56
56 gm dissolved in d/w to make final volume 1 liter.
i.e., 56 gm in 1,000 ml = 1 mol/lt
5.6 gm in 100 ml = 1 mol/lt
0.56 gm in 100 ml = 0.1 mol/lt

Hydrochloric acid concentrated
Density = 1.19
Mol. Wt. = 36.5
Assay = 36%
D=M/V
V= M/D = 36.5/1.19 ml required for 100%
36.5*100/1.19 ml required for 1%
36.5*100/1.19*36 ml required for 36%
= 85.2 ml
85.2 ml HCl mixed with d/w to make final volume 1 liter
i.e.,85.2 ml in 1,000 ml d/w = 1 mol/lt
8.52 ml in 100 ml d/w = 1 mol/lt
0.852 ml in 100 ml d/w = 0.1 mol/lt

Sulphuric acid
Density = 1.84
Assay = 95%
Mol. Wt. = 98
V=M/D = 98/1.84 ml required for 100%
98*100/1.84*95 ml required for 95%
= 56.06 ml
56.06 ml H2SO4 mixed with d/w to make final volume 1 liter
i.e.,56.06 ml in 1,000 ml d/w = 1 mol/lt
5.606 ml in 100 ml d/w = 1 mol/lt
0.5606 ml in 100 ml d/w = 0.1 mol/lt
2.803 ml in 500 ml d/w = 0.1 mol/lt

NaOH
Assay = 97%
Mol. Wt. = 40 gm
40 gm required for 100%
40*100/97 required for 97%
41.23 gm NaOH mixed in d/w to make final volume 1 liter
i.e., 41.23 gm in 1,000 ml d/w = 1 mol/lt
4.123 gm in 100 ml d/w = 1 mol/lt
12.369 gm in 300 ml d/w = 1 mol/lt
1.2369 gm in 300 ml d/w = 0.1 mol/lt

NaH2PO4
Mol. Wt. = 156
156 gm mixed in d/w to make final volume 1 liter
i.e., 156 gm in 1,000 ml d/w = 1 mol/lt
1.56 gm in 100 ml d/w = 0.1 mol/lt
3.12 gm in 100 ml d/w = 0.2 mol/lt

Acetic acid
Specific gravity = 1.047
Mol.wt. = 60.05
V=M/D = 60.05/1.047 ml in d/w to make final volume 1 liter
= 57.35 ml
i.e., 57.35 ml in 1,000 ml d/w = 1 mol/lt
5.735 ml in 100 ml d/w = 1 mol/lt
0.5735 ml in 100 ml d/w = 0.1 mol/lt
1.147 ml in 100 ml d/w = 0.2 mol/lt

1.147 ml acetic acid mixed with d/w to make final volume 100 ml in volumetric flask


I hope it will be helpful for all those lost in confusion. Let me know whether it helped you or not. Thank you