calculate the molarity of sodium chloride in a half-normal saline solution (0.45% nacl). the molar mass of nacl is
58.44 g/mol.

The molarity of half-normal saline solution (0.45% NaCl) is gotten to be; 0.077 M.

We are given;

The concentration of half normal saline solution(NaCl) = 0.45% = 0.45g/100g

From conversion, we know that 100 g = 100 mL

Thus converting to liters;

Volume of Solution = 100/1000 L

Volume of Solution = 0.1 L

Now,

The molar mass of NaCl = 58.44 g/mol

Th Formula to calculate Molarity is given as;

M = Number of moles of solute/Volume of solution

We don't have number of moles here.

Thus, since concentration is 0.45g/100g, and molar mass of NaCl = 58.44 g/mol, then;

Number of moles of NaCl =   0.45g/58.44 g/mol

Number of moles of NaCl = 0.0077 mol

Thus, Molarity is;

M = 0.0077mol/0.1 L

M = 0.077 M

In conclusion, the molarity of half-normal saline solution (0.45% NaCl) is gotten to be 0.077 M.

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0.077 M

Explanation:

Data Given :

The concentration of half normal (NaCl) saline = 0.45g / 100 g

So,

Volume of Solution = 100 g = 100 mL

Volume of Solution in Liter = 100 mL / 1000

Volume of Solution = 0.1 L

molar mass of NaCl = 58.44 g/mol

Molarity:

Molarity is the representation of the solution. It is amount of solute in moles per liter of solution and represented by M

Formula used for Molarity

                M = moles of solute / Liter of solution . . . . . . . . . . (1)

Now to find number of moles of Nacl

                no. of moles of NaCl = mass of NaCl / molar mass

                no. of moles of NaCl = 0.45g / 58.44 g/mol

               no. of moles of NaCl = 0.0077 g

Put values in the eq (1)

                  M = moles of solute / Liter of solution . . . . . . . . . . (1)

                  M = 0.0077 g / 0.1 L

                  M = 0.077 M

So the molarity of half-normal saline solution (0.45% NaCl) = 0.077 M

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