How many grams of nacl are present in 350 ml of 0.020 m nacl solution? (remember that 1 mole nacl = 58.44 g/mole.)
There are a number of
ways to express concentration of a solution. This includes molarity. Molarity
is expressed as the number of moles of solute per volume of the solution. The
mass of NaCl needed for the solution is calculated as follows:
0.020 M = 0.020 moles NaCl / L
amount in moles of solute = 0.020 moles NaCl / L x .350 L = 0.007 mol NaCl
amount in grams =0.007 mol NaCl x 58.44 g/mol = 0.41 g NaCl
Answer : The mass of NaCl present in solution are 0.41 grams.
Explanation : Given,
Volume of solution = 350 mL
Molar mass of NaCl = 58.44 g/mole
Molarity of NaCl = 0.020 M
Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.
Formula used :
\(\text{Molarity}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}\)
In this question, the solute is NaCl.
Now put all the given values in this formula, we get:
\(0.020M=\frac{\text{Mass of solute}\times 1000}{58.44g/mole\times 350mL}\)
\(\text{Mass of solute}=0.41g\)
Therefore, the mass of NaCl present in solution are 0.41 grams.