What is the molarity of a 2.4-liter solution containing 124 grams of HF?

Question

Caleb Adams · Accepted Answer

#"2.6 M"# Molar mass of #"HF"# is #"20 g/mol"# Moles of #"HF"# in solution is #"n" = "124 g"/"20 g/mol" = "6.2 mol"# #"Molarity" = "Moles of solute"/"Volume of solution (in litres)"# #"Molarity" = "6.2 mol"/"2.4 L" = "2.6 M"#

Jeremiah Adams · Answer

Well, what is #"molarity"#?

By definition, #"molarity"="moles of solute"/"volume of solution"#. And so we take the quotient...#((124*g)/(20.01*g*mol^-1))/(2.40*L)=2.58*mol*L^-1#...are the units of concentration correct?

Ellie Andrews · Answer

undefined To find the molarity of the solution, you need to first calculate the number of moles of HF present in the solution using its molar mass, then divide this by the volume of the solution in liters.

1. Calculate the number of moles of HF:
$$ \text{Molar mass of HF} = 1 \times \text{atomic mass of H} + 1 \times \text{atomic mass of F} $$
$$ = 1 \times 1.008 \, \text{g/mol} + 1 \times 18.998 \, \text{g/mol} $$
$$ = 1.008 \, \text{g/mol} + 18.998 \, \text{g/mol} $$
$$ = 20.006 \, \text{g/mol} $$

$$ \text{Number of moles of HF} = \frac{\text{mass of HF}}{\text{molar mass of HF}} $$
$$ = \frac{124 \, \text{g}}{20.006 \, \text{g/mol}} $$
$$ = 6.197 \, \text{mol} $$

2. Calculate the molarity:
$$ \text{Molarity} = \frac{\text{number of moles}}{\text{volume of solution in liters}} $$
$$ = \frac{6.197 \, \text{mol}}{2.4 \, \text{L}} $$
$$ = 2.582 \, \text{M} $$

The molarity of the solution is $ 2.582 \, \text{M} $.