Stoichiometry and percentage concentration [closed]

Question

10 g of Mg(NO3)2 was dissolved in 120 g of water. The result is a solution with a density of 1.12 g / ml. Determine the percentage concentration (in %) and the molar concentration (in mol/L) of the resulting solution

Answer 1

As you are new to this forum I want to welcome you but nevertheless I would kindly ask you to first make your own research on a problem you have and then ask your question providing some information on what you do not understand. Your question seems to be a homework which you only copied without any signs of your own thoughts.

That said, here are some explanations regarding your assignment.

Question 1

If you put 10 grams of something into 120 grams of something else, than your resulting mass is 130 grams, right? So, 10 g of those 130 g are how many percent?

$$ \frac{10\,\text{g}}{130\,\text{g}} \times 100\,\text{%} = 7.69 \% $$

Question 2

Magnesium nitrate ($\ce{Mg(NO3)2}$) has a molecular weight of 148.31 g/mol and you have used 10 g, so what is the amount of magnesium nitrate?

$$ \begin{align} M &= \frac{m}{n}\\ \Rightarrow n &= \frac{m}{M}\\ &= \frac{10\,\text{g}}{148.31\,\text{g/mol}}\\ &= 6.74 \times 10^{-2}\,\text{mol} \end{align} $$

As you know, water has a density of 1 gram per milliliter, thus 120 grams correspond to 120 milliliters, which, again, is equal to 0.12 liters.

To calculate the concentration in mol per liter, just divide the amount in mol by the volume in liter:

$$ \begin{align} c &= \frac{n}{V}\\ &= \frac{6.74 \times 10^{-2}\,\text{mol}}{0.12\,\text{L}}\\ &= 5.62 \times 10^{-1}\,\frac{\text{mol}}{\text{L}} \end{align} $$

Check

Let's check if our results are correct:

If you happen to have 120 milliliters of a 0.562 molar solution, then you can easily calculate the amount of magnesium nitrate by multiplication:

$$ \begin{align} n &= c \times V\\ \Rightarrow n &= 5.62 \times 10^{-1}\,\frac{\text{mol}}{\text{L}} \times 0.12\,\text{L}\\ &= 6.74 \times 10^{-2}\,\text{mol} \end{align} $$

If magnesium nitrate has a molecular mass of 148.31 g/mol, then the mass can, again, be easily calculated by multiplication:

$$ \begin{align} m &= M \times n\\ \Rightarrow m &= 148.31\,\text{g/mol} \times 6.74 \times 10^{-2}\,\text{mol}\\ &= 10\,\text{g} \end{align} $$