Question

Ile gram kwasu octowego CH3COOH znajduje się w 3 dm szesciennych  4 molowego roztworu?
proszę o obliczenie.

Answer 1

Dane:                                        Cm = n/v
v = 3 dm³                                  n = Cm * v
Cm = 4mole/dm³                       n = 4 mole/dm³ * 3 dm³
MCH3COOH = ?                         n = 12 moli


MCH3COOH = 2*12 + 4*1 + 2*16 = 60 g/mol

                                             1 mol - 60 g/mol
                                             12 moli - MCH3COOH
             
                                          12 moli * 60 g/mol
                  MCH3COOH = ---------------------------
                                                1 mol
                   
                       MCH3COOH = 720 g/mol