Milliliters of NaBr when changing molarity [closed]

Question

I was very confused by this problem. I was able to get the molarity, but I wasn’t certain how to go from there:

If $\pu{2.60 g}$ of $\ce{NaBr}$ is dissolved in enough water to make $\pu{160.0 mL}$ of solution, what is the molar concentration of $\ce{NaBr}$? How many milliliters of $\pu{0.10 M}$ $\ce{NaBr}$ would you need to supply $\pu{2.60 g}$ of $\ce{NaBr}$?

$\text{Molarity} = \pu{0.158 moles/L}$

I wasn’t certain how to answer the second part of the problem. I kept getting $\pu{159 mL}$, but the answer is $\pu{250 mL}$. What am I doing wrong?

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Edit:

I am not entirely used to the M symbol for molarity, so I was mixing myself up. Instead, I equated $\pu{0.10 mol/L} = \frac{(\pu{2.60 g} \times \text{molar mass NaBr})}{x \ \pu{L}}$ — I solved from there for $x$ and got $\pu{250 mL}$.

Answer 1

I decided to give a help here since OP has given enough effort to solve this problem. First, OP needs to know the molarity of a solution is defined as the amount of solute (in this case $\ce{NaBr}$) in $\pu{mol}$ per $\pu{1.0 L}$ of solution. The first part of problem states:

If $\pu{2.60 g}$ of $\ce{NaBr}$ is dissolved in enough water to make $\pu{160.0 mL}$ of solution, what is the molar concentration of $\ce{NaBr}$?

To solve this problem you may need to know the molar mass of $\ce{NaBr}$, which is not given. But you can find it by googling it. The molar mass of $\ce{NaBr}$ is found as $\pu{102.8 g mol-1}$. Now you can find the amount of $\ce{NaBr}$ in $\pu{160.0 mL}$ or $\pu{0.160 L}$ of solution (volume has to be in $\pu{L}$), which is:

$$\frac{\pu{2.60 g}}{\pu{102.8 g mol-1}} = \pu{0.0253 mol}$$

Thus,

$$\text{Molarity of the solution} = \frac{\text{amount of NaBr}}{\text{volume of solution}} = \frac{\pu{0.0253 mol}}{\pu{0.160 L}} = \pu{0.158 mol L-1}\\ = \pu{0.158 M}$$

The second part of the question is:

How many milliliters of $\pu{0.10 M}$ $\ce{NaBr}$ would you need to supply $\pu{2.60 g}$ of $\ce{NaBr}$?

This supposed to be solved using the first part of the reaction. So, by first part, you know now that $\pu{0.160 L}$ of $\pu{0.158 M}$ $\ce{NaBr}$ contains exactly $\pu{2.60 g}$ of $\ce{NaBr}$. so using the dilution principle, $M_1V_1 = M_2V_2$, you can find your unknown:

Here, $M_1 = \pu{0.158 M}$, $V_1 = \pu{0.160 L}$, and $M_2 = \pu{0.10 M}$. Your unknown is $V_2$, you can find using $M_1V_1 = M_2V_2$:

$$V_1 = \frac{M_1V_1}{M_2} = \frac{\pu{0.158 M} \times \pu{0.160 L}}{\pu{0.10 M}} = \pu{0.253 L} = \pu{253 mL} $$

Since olarity of the solution is given as $\pu{0.10 M}$, the answer should be $\pu{250 mL} $ with correct significant figures.

Note: @Entropy has shown the correct calculation as you have done. However, it is a different method to find the volume of solution. I'm sure the one wrote the question intended to use the first part of the question to solve the second part.

Answer 2

It comes to be $\pu{252 mL}$ precisely:

$$V = \frac{\pu{2.6 g}}{\pu{103 g}} \times \frac{1}{0.1} \times \pu{1000 mL} = \pu{252.4 mL}$$

$\text{Molar Mass of NaBr} = \pu{103 g mol-1}$.