Answer:
93.6 g of NaCl.
Explanation:
We'll begin by converting 800 mL to litre (L). This can be obtained as follow:
1000 mL = 1 L
Therefore,
800 mL = 800 mL × 1 L / 1000 mL
800 mL = 0.8 L
Next, we shall determine the number of mole of NaCl in the solution. This can be obtained as follow:
Molarity = 2 M
Volume = 0.8 L
Mole of NaCl =?
Molarity = mole / Volume
2 = mole of NaCl / 0.8
Cross multiply
Mole of NaCl = 2 × 0.8
Mole of NaCl = 1.6 moles
Finally, we shall determine the mass of NaCl needed to prepare the solution. This can be obtained as follow:
Mole of NaCl = 1.6 moles
Molar mass of NaCl = 23 + 35.5
= 58.5 g/mol
Mass of NaCl =?
Mole = mass /Molar mass
1.6 = Mass of NaCl / 58.5
Cross multiply
Mass of NaCl = 1.6 × 58.5
Mass of NaCl = 93.6 g
Therefore, 93.6 g of NaCl is needed.
93,6 grams of NaCl is needed to prepare a 800ml of 2M sodium chloride.
In a 2M solution, there are 2 moles of the substance per 1 liter, that is, in 800 ml we have:
[tex]\frac{2 mol}{x mol} =\frac{1000 ml}{800ml}[/tex]
[tex]x = 1.6 mol[/tex]
Thus, 1.6 mol of NaCl is needed to prepare the solution.
To calculate the mass in an amount of moles, we use the molar mass, so that:
[tex]MM_{NaCl}= 58.5g/mol[/tex]
[tex]m = MM \times mol[/tex]
[tex]m = 58.5 \times 1.6[/tex]
[tex]m = 93.6g[/tex]
So, 93,6 grams of NaCl is needed to prepare a 800ml of 2M sodium chloride.
Learn more about mole calculation in: brainly.com/question/2845237
