1. How to prepare 15ml of 20% w/v NaCl solution?
2. How many grams of NaCl are required to prepare 45g of NaCl solution (35%
w/w)?
3. Calculate the molarity of 0.289 moles of FeCl3 dissolved in 120 ml of
solution.
4. How many moles of NaCl are present in 600 ml of a 1.55 M NaCl solution?
5. What is the molarity of 650 ml of solution containing 63 grams of NaCl?
6. How many grams of CaCO3 are needed to produce 500 ml of 1.66 M CaCO3
solution? How to prepare this solution?
7. What volume of a 0.88 M solution can be made using 130 grams of FeCl2?
8. Prepare 70ml of 1.5M NaCl solution from a 3M solution.
9. What is the volume of a 2 M H2SO4 solution prepared using 3 ml of 6 M
H2SO4 solution?
10.What is the molarity of a 17 ml solution used to prepare 65 ml of 2.4 M
solution

Weight by volume is a concentration determining the method of the solution. 20 % w/v NaCl is prepared by mixing 3 gm in 15 mL.

What is concentration?

Concentration is the amount of the solute or the solvent present in the solution. It can be calculated by molarity, molality, w/v, etc.

If the solution is 20% w/v then,

( W ÷ 15 ) × 100=20

W = (20 × 15) ÷ 100 =3g

Hence, 3 gm NaCl must be added to 15 mL solution to make it 20% w/v.

Grams of NaCl for 35 % w/v is calculated as:

( w ÷ 45) × 100=35

w = (35 × 45) ÷ 100

= 15.75 grams

Hence, 15.75 gm NaCl is required to prepare 45 gm solution.

Molarity is calculated as:

Molarity = number of moles ÷ volume in litres

Given,

Moles = 0.289

Volume = 0.12L

Molarity = 0.289 ÷ 0.12 = 2.4M

Hence, 2.4 M is the molarity of 120 mL of ferric chloride solution.

Moles of sodium chloride is calculated as:

Number of moles = Molarity × Volume in litres

Given,

Volume = 0.6 L

Molarity = 1.55 M

Substituting values:

Number of moles = 1.55 × 0.6 = 0.93Moles

Hence, 0.93 moles are present in 1.55 M solution.

Molarity of 63 gm NaCl is calculated as:

Given,

Mass = 63 gm

Volume = 0.65 L

Molarity = mass ÷ (molar mass × volume)

M = 63 ÷ ( 58.44 × 0.65)

= 63 ÷ 37.98

= 1.66 M

Hence, 1.66 M is the molarity.

Mass of calcium carbonate from molarity is calculated as:

Given,

Molarity = 1.66 M

Volume = 0.5 L

Mass of calcium carbonate = Molarity × molar mass × volume

1.66 × 100.0869 × 0.5 = 83.07 gm

Hence, 83.1 gm is mass required of calcium carbonate to make 1.66 M solution.

Volume of solution from molarity is calculated as:

Given,

Mass of ferrous chloride = 130 gm

Molar mass of ferrous chloride = 126.751

Molarity = 0.88 M

Volume = mass ÷ (molar mass × molarity)

= 130 ÷ (126.751 × 0.88)

= 1.16 mL

Hence, 1.16 mL solution can be prepared.

The dilution formula is used to calculate the initial volume as:

M1 = 3 M

M2 = 1.5 M

V2 = 70 mL

V1 is calculated as:

M1V1 = M2V2

V1 = (1.5 × 70) ÷ 3

= 35 mL

Hence, 35 mL of stock solution is required to prepare the 1.5 M solution.

The final volume to prepare the solution is calculated as:

M1 = 6 M

V1 = 3 mL

M2 = 2 M

V2 is calculated as:

M1V1 = M2V2

(6 × 3) ÷ 2 = V2

= 9 ml

Hence, 9 mL is the volume of the desired solution.

The molarity of 17 mL solution is calculated as:

Given,

V1 = 17 mL

M2 = 2.4 m

V2 = 65 mL

M1V1 = M2V2

M1 = (2.4 × 65) ÷ 17

= 9.1 M

Hence, 9.1 M is the molarity of the solution that is used to prepare 2.4 M solution.

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