Question

1. How many mL of 5.75 M HCl do you need to make up 2.50 L of 0.235 M HCl?

2. If you start with 2.94 g of KNO3 (s) and dissolve it in 100.0 g of water what is the mass percent of KNO3 (aq) solution?

3. If you dilute 18.2 mL of 1.05 M NaOH up to 0.500 L whats the new concentration in M?

4. How much water needs to be added to 35.00 g of glucose to make a 2.500 % solution by mass?

Answer 1

1.

The formula which is needed is:

V₁S₁ = V₂S₂

Where, V = Volume, S = Strength, and 1 and 2 denote initial and final solutions respectively.

While using this formula, the units of Volume and Strength on both sides of the equation must be same.

For the given question, the final solution required is 2.50 L of 0.235 M HCl. So,

V₂ = 2.50 L, S₂ = 0.235 M

The initial solution taken has a strength (S₁) of 5.75 M

We need to find volume of initial solution needed, that is V₁

Applying the formula :

V₁ x 5.75 M = 2.50 L x 0.235 M

Solving, we get:

V₁ = 0.102 L = 102 mL ( 1 L = 1000 mL)

2.

Formula :

Here, mass of component KNO₃ = 2.94 g

Total mass of sample = mass of KNO₃ + Mass of water = 2.94 g + 100.0 g = 102.94 g

Putting in formula, mass % of KNO₃ = = 2.86 %

3.

V₁S₁ = V₂S₂

Initial volume ( V₁) = 18.2 mL, initial strength ( S₁) = 1.05 M

Final volume = 0.500 L = 500 mL

Putting in formula :

(18.2 mL)( 1.05 M)= (500 mL) S₂

So,S₂= final concentration = 0.03822 M

4.

Mass of component (glucose) = 35.00 g

Total mass of solution = mass of glucose + water = 35.00 g + mass of water

Mass % = 2.500 %

Solving,mass of water = 1365 g