Question

How many grams of sodium hydroxide are needed to make 250. Ml of a 7.80 m solution?

Answer 1

Answer;

= 78 g of NaOH

Explanation;

Concentration = Moles of solute/Volume of solution

Thus;

Moles of the solute = Volume × Concentration

                                = 7.80 Moles/L × 0.250 L

                                 = 1.95 moles

But; 1 mole of NaOH = 40.0 g

Thus;

Mass of NaOH = moles × molar mass

                         = 1.95 moles × 40 g/mole

                        = 78 g of NaOH

Answer 2

78 g of NaOH

Further explanation

Given:

• The concentration of the NaOH solution that we want to make is 7.8 M.
• The volume of NaOH solution that we want to make is 250 ml.

Question:

How many grams of sodium hydroxide are needed to make 250 ml of a 7.80 M solution?

The Process:

Step-1

Let us prepare the relative molecular mass (Mr) of NaOH.

• Relative atomic mass: Na = 23, O = 16, and H = 1
• Relative molecular mass (Mr) of NaOH = 23 + 16 + 1 = 40.

Step-2

The concentration of the solution is a mole of solute divided by 1 liter of solution.

$\boxed{ \ M = \frac{n}{V} \ }$  (Remember V in liters here)

We know that mole is mass divided by molecular mass.

$\boxed{ \ n = \frac{mass}{Mr} \ }$

The relationship between concentration (M), mass, relative molecular mass (Mr), and volume of solution in ml (V) is as follows:

$\boxed{ \ M = \frac{mass}{Mr} \times \frac{1,000}{V} \ }$ (Remember V in ml here)

Substitute all known data into the equation.

$\boxed{ \ 7.80 = \frac{mass}{40} \times \frac{1,000}{250} \ }$

$\boxed{ \ 7.80 = \frac{mass}{40} \times 4 \ }$

$\boxed{ \ 7.80 = \frac{mass}{10} \ }$

Mass = 7.80 x 10

Mass = 78

Thus, the number of grams of sodium hydroxide are needed to make 250 ml of a 7.80 M solution is 78 grams.

_ _ _ _ _ _ _ _ _ _

Notes:

The concentration of the solution, especially in this case molarity, represents one significant data. Among them play a role in

• determining the pH value (acid, base, or salt),
• determining the value of the equilibrium constant, and
• calculating osmotic pressure.

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