Question

a) 0.250 mol of Na2S in 1.10 L of solution.

b) 20.9 g of MgS in 831 mL of solution.

c) You have 47.0 mL of a 0.400 M stock solution that must be diluted to 0.100 M. Assuming the volumes are additive, how much water should you add?

EXPLAIN PLEASE.

Answer 1

Calculating Molarity

a) 0.250 moles Na2S/1.10 L= 0.250/1.10 = 0.23 M

b) 20.9 g of MgS in 831 mL of solution

molar mass of MgS = 24.3+32.1=56.4 g

so 20.9g =20.9/56.4=0.37 mole in 831ml or 0.831L

so molarity number of moles by liter = 0.37/0.831=0.445 M

C) You have 47.0 mL of a 0.400 M stock solution that must be diluted to 0.100 M

M1V1 = M2V2

M1= 0.400 M
V1 = 47.0 mL
M2 = 0.100 M
V2 = unknown (what we will solve for; this will be the TOTAL volume)

V2 = (M1V1)/M2 = ((0.400 M)*(47.0 mL))/(0.100 M)

V2 = 188 mL

The water needed is the difference between the total volume (V2) and V1:

volume of water = V2 - V1 = 188 mL - 47 mL = 141 mL