Question

What mass of H3PO4 (98.0 g/mol) is present in 86.3 L of a 0.0823 M solution of H3PO4?

Select one:
a. 10.7 g
b. 6.96 x 102 g
c. 0.00143 g
d. 1.03 x 105 g
e. 0.0724 g

Answer 1

Start by using C=n/v

-> n = c*v
-> n = 86.3*.0823M
-> n = 7.14209 mol

Then find mass by using n = m/M
-> m = n*M
-> m = 7.14209*98
-> m = 699.92 g H3PO4

Sigfigs -> 700 g H3PO4

Hope this helps