How can you calculate the volume of .270 M solution that can be prepared from 70.5g of K2CO3?

Answer 1

Using the formula,

Molarity = No. of moles / Volume of solution (in litres).......eq.1

First, we'll calculate the no. of moles,

Molar mass of K2CO3 = (2 x 39) + (12) + (3 x16) = 138 g
Mass of K2CO3 given = 70.5 g

Hence, no. of moles = given mass/molar mass = 70.5/138 = 0.51 moles

To go back to eq. 1,

molarity is given as .270 M

Hence, volume of solution =
no. of moles/molarity = 0.51/0.270 = 1.89 L

Answer 2

M = moles / liter of solution

Change 70.5 g K2CO3 to moles:

70.5 g / 138 molar mass = 0.51 mol K2CO3

0.270 M = 0.51 mol / ? L

0.51 mol / 0.270 M = L = 1.89 L

Answer 3

K2CO3= 138g/mol

70.5g/138g/mol=0.511 mol of K2CO3

Molarity= mol/volume thus volume=mol/molarity
=.511/0270
= 1.89 L or 1890 mL