What is the molarity of a solution made by dissolving 20.0 g of #H_3PO_4# in 50.0 mL of solution?

Question

Henry Antrim · Accepted Answer

#"Molarity"# #=# #"Moles of solute"/"Volume of solution"# So #"concentration"# #=# #(20.0*g)/(98.08*g*mol^-1)xx1/(50.0xx10^-3L)# #~=# #4*mol*L^-1# w.r.t. #H_3PO_4#.

Olivia Anton · Answer

undefined To find the molarity of the solution, we first need to calculate the number of moles of H3PO4, then divide by the volume of the solution in liters.

1. Calculate the molar mass of H3PO4:
   $1 \times 3(1.01) + 1 \times 1(15.999) + 4 \times 1(30.974) = 98.00 \, \text{g/mol}$

2. Convert the mass of H3PO4 to moles:
   $20.0 \, \text{g} \times \frac{1 \, \text{mol}}{98.00 \, \text{g}} = 0.204 \, \text{mol}$

3. Convert the volume of the solution to liters:
   $50.0 \, \text{mL} \times \frac{1 \, \text{L}}{1000 \, \text{mL}} = 0.0500 \, \text{L}$

4. Calculate the molarity ($M$):
   $M = \frac{\text{moles of solute}}{\text{volume of solution (in liters)}}$
   $M = \frac{0.204 \, \text{mol}}{0.0500 \, \text{L}} = 4.08 \, \text{M}$

Therefore, the molarity of the solution is 4.08 M.