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What mass of solute is needed to prepare each of the following solutions? a. 3.00 L of 0.125 M K2SO4 b. 550 mL of 0.015 M NaF c. 700 mL of 0.350 M C6H12O6

Verified Answer :

Edufirst

a.
M = 0.125 M
V = 3.0 L
M=n/V(L) ⇒ n = 0.125 mol/L 3.0L = 0.375 mol

MM of K2SO4

K2 = 2 39 = 78 g/mol
S= 32 g/mol
O4 = 416 = 64 g/mol

MM = 78g/mol + 32g/mol + 64g/mol =174 g/mol

mass = nMM = 0.375 mol174 g/mol = 65.25 g

b.
M= 0.015 M
V = 0.55L

n = 0.015 mol/L 0.55L = 0.00825 mol

MM NaF

Na: 23g/mol
F: 19 g/mol

MM = 23g/mol + 19g/mol = 42g/mol

mass = n MM = 0.00825 mol 42g/mol = 0.3465 g

c.

n = MV(L) = 0.350 mol/L 0,700L = 0.245 mol
MM = 612 g/mol + 121g/mol + 616g/mol = 180 g/mol

mass = nMM = 0.245 mol * 180g/mol = 44.1 g
 

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