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How many grams of sodium chloride are required to prepare 500.0 ml of a 0.100 m solution?
a.1.46 g strike reset
b.2.93 g strike reset
c.29.3 g strike reset
d.58.5 g strike reset?

Respuesta :

molarity is the number of moles of solute in 1 L solution.
molarity of solution to be prepared is 0.100 M
volume of solution to be prepared is 500.0 mL 
the number of moles in 1 L - 0.100 mol 
therefore number of moles in 500.0 mL - 0.100 mol/L x  500.0 x 10⁻³ L = 0.0500 mol
mass of NaCl required - 0.0500 mol x 58.5 g/mol = 2.925 g
mass of NaCl needed is therefore 2.93 g
answer is B. 2.93 g
dsdrajlin

In order to prepare 500.0 mL of a 0.100 M NaCl solution, 2.93 g of sodium chloride are required (Option b).

What is molarity?

Molarity is the number of moles of solute per liters of solution.

We have 500.0 mL of a 0.100 M NaCl solution. First, we will calculate the moles of solute using the definition of molarity.

M = moles solute / liters solution

moles solute = M × liters solution

moles solute = 0.100 mol/L × 0.5000 L = 0.0500 mol

Next, we will convert 0.0500 moles of NaCl to grams using its molar mass.

0.0500 mol × 58.5 g/mol = 2.93 g

In order to prepare 500.0 mL of a 0.100 M NaCl solution, 2.93 g of sodium chloride are required (Option b).

Learn more about molarity here: https://brainly.com/question/26873446

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