What mass in g of NaCl is required to prepare 250.0 mL of a 0.3210 M NaCl solution? mass NaCl: Answer g Describe how to prepare this solution using a 250.0 mL volumetric flask by placing the steps in the correct order. Not all of the steps will be used.

Explanation:

Molarity = 0.3210M

Volume = 250ml

Mass = ?

Molarity is defined as the number of moles of solute per liter of solution. It's unit is mol/L.

This means 0.3210 solution contains  0.3210 moles of sodium chloride in 1.0 L of solution.

The target solution has a volume of 250ml. Upon conversion to L by dividing by 1000 we have;

250 / 1000 = 0.250L

if 0.3210 moles = 1L

0.250 = x

x = 0.250 * 1 / 0.3210

x = 0.7788 moles

Sodium chloride has a molar mass of 58.44 g mol−1, which basically means that one mole of sodium chloride has a mass of 58.44 g.

1 mole = 58.44g

0.7788 mole = x

9

x = 0.7788 * 58.44 / 1

x = 45.51g

To prepare 0.3210 M NaCl; dissolve (45.51g / 4 = 11.3775g) of Nacl in (1 L / 4 = 250ml) of water.

the grams   of   barium that   are in   2.3 x 10^23   atoms of barium   is   52.4 g (answer c)

    calculation

mass = moles   of ba   x molar   mass of ba

  molar mass of ba from periodic table=   137.3 g/mol

  use the   avogadro's law   constant   to   calculate the moles of barium

  that   is     1 mole =   6.02   x 10^23 atoms

                    ? moles = 2.3 x 10^23   atoms

by cross   multiplication

=(1 mole   x 2.3x 10^23 atoms)   / 6.02   x10^23 atoms)=   0.382   moles

mass   is therefore = 137.3   g/mol x 0.382 mol = 52.4 ( answer c)

explanation: