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2 years ago

How many mL of a 4.50 (m/v)% solution would contain 23.0 g of solute?

Chemistry

2 answers:

nlexa [21]2 years ago

4 0

23.0 g/ (4.50 g/ mL) gives 5.11 mL as the volume.

Helen [10]2 years ago

3 0

<u>Answer:</u> The given amount of solute will be present in 511.11 mL of solution.

<u>Explanation:</u>

We are given:

Concentration of solution = 4.50 (m/v) %

This means that 4.50 grams of solute is present is 100 mL of solution.

We need to find the volume of solution that can contain 23.0 grams of solute.

By applying unitary method, we get:

4.50 grams of solute is present in 100 mL of solution.

So, 23.0 grams of solute will be present in = $\frac{100mL}{4.50g}\times 23.0=511.11mL$ of solution.

Hence, the given amount of solute will be present in 511.11 mL of solution.

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2 years ago

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2 years ago

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salantis [7]

Answer:

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Explanation:

Hello there!

In this case, according to the given chemical reaction:

2 Al + 3 Cl2 --> 2 AlCl3

Whereas there is a 2:3 mole ratio of aluminum to chlorine; it will be possible for us to calculate the required grams of aluminum by using the equality 22.4 L = 1 mol, the aforementioned mole ratio and the atomic mass of aluminum (27.0 g/mol) to obtain:

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