Answer :
164.22 grams of sodium fluoride are dissolved in 3.75 L of a 1.55 M solution.
To find the number of grams of sodium fluoride (NaF) dissolved in 3.75 L of a 1.55 M solution, you can use the formula:
molarity (M) x volume (L) x molar mass (g/mol) = mass (g)
Sodium fluoride (NaF) has the following molar mass:
22.99 g/mol (Na) + 19.00 g/mol (F) = 41.99 g/mol
So the mass of sodium fluoride dissolved in 3.75 L of a 1.55 M solution is:
mass (g) = 1.55 M x 3.75 L x 41.99 g/mol = 164.22 g
Therefore, 164.22 grams of sodium fluoride are dissolved in 3.75 L of a 1.55 M solution.
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