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2 years ago

35.0 mL of 12.0 M HCl is added to enough water to have a final volume of 1.20 L. What is the molarity of the final solution?

Chemistry

1 answer:

creativ13 [48]2 years ago

6 0

<h3>Answer:</h3>

0.35 M

<h3>Explanation:</h3>

<u>We are given;</u>

Initial volume as 35.0 mL or 0.035 L
Initial molarity as 12.0 M
Final volume is 1.20 L

We are required to determine the final molarity of the solution;

Dilution involves adding solvent to a solution to make it more dilute which reduces the concentration and increases the solvent while maintaining solute constant.

Using dilution formula we can determine the final molarity.

M1V1 = M2V2

Rearranging the formula;

M2 = M1V1 ÷ V2

= (12.0 M × 0.035 L) ÷ 1.2 L

= 0.35 M

Thus, the final concentration of the solution is 0.35 M

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A solution of NaCl has a concentration of 1.50 M. The density of the solution is 1.06 g/mL. What is the concentration solution b

gavmur [86]

<u>Answer:</u>

<em>The concentration solution by percent in mass is 8.28%</em>

<u>Explanation:</u>

Molarity is moles of solute present in 1L of its solution .

Hence the formula is

$\left.\text {Molarity}=\frac{\text {moles of solute}}{\text {Volume of solution in } L} \text { (unit is } \frac{\text { mol }}{L} \text { or } M\right)=\frac{1.50 \mathrm{mol}}{1 \mathrm{L}}$

Let us convert the numerator and the denominator into mass in g

Numerator :

$mass $=$ moles $\times$ molar mass\\\\$=1.50 \mathrm{mol} \times 58.5 \frac{g}{\text { mol }}=87.75 \mathrm{g}$ (solute)$

Denominator :

$\begin{aligned} \text {mass} &=\text { density } \times \text {volume}=\frac{1.06 \mathrm{g}}{1 \mathrm{ml}} \times 1 \mathrm{L} \\\\ &=\frac{1.06 \mathrm{g}}{1 \mathrm{ml}} \times 1000 \mathrm{ml}=1060 \mathrm{g} \end{aligned}$

The formula to find mass percentage is

$\%$ mass $=\frac{\text { mass of solute }}{\text { mass of solution }} \times 100 \%$

$\%$ mass of $\mathrm{NaCl}=\frac{\text { mass of solute }}{\text { mass of solution }} \times 100 \%$

$$=\frac{87.75 g}{1060 g} \times 100 \%$\\$

=8.28% (Answer)

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