Currently I am attempting two questions on concentration but I am not sure if I am correct. Can someone help me with these questions? These are the questions:
1) Calculate the volume of pure water added to dilute $\pu{0.2 M}$ $\ce{HCl}$ solution in $\pu{250 cm3}$ to $\pu{0.12 M}$ $\ce{HCl}$ solution.
$\pu{0.2 M}$ solution contains $\pu{0.05 mol}$. $\pu{0.12 M}$ solution contains $\pu{0.03 mol}$. The difference between the two solutions is $\pu{0.02 mol}$ (Which is how much water was added), so
$$V(\ce{H2O}) = 0.02 \times (2 \times 1 + 1 \times 16) = \pu{0.36 L}$$
2) Calculate the mass of $\ce{Na}$ that will form $\pu{0.15 M}$ solution of $\ce{NaOH}$ by reacting with $\pu{400 cm3}$ of water.
I know that this method is totally wrong but I'll show you:
$$\pu{0.15 M} = \frac{n}{0.4} \implies n = 0.06$$ $$m(\ce{Na}) = 0.06 \times 23 = \pu{1.38 g}$$
where $n$ just refers to the concentration ($C = n/V$).
Edit:
Thanks to @George Tian I have made another attempt for the second question.
I have calculated the number of moles of NaOH for a 0.15M solution:
c = n/v
n = 0.4 x 0.15
= 0.06
And I calculated how much Na is needed for this reaction:
The balanced equation is: 2 Na + 2 H20 -> 2 NaOH + H2
So, 0.06 mol of Na is needed (Due to the molar ratio)
Then I converted moles to grams
Mass = 0.06 x 23
= 1.38 g