Question

Calculate the molarity of each of the following solutions.

A) 0.36 mol of LiNO3 in 6.24 L of solution

B) 74.0 g C2H6O in 2.43 L of solution

C) 12.43 mg KI in 110.3 mL of solution

Answer 1

A)

volume , V = 6.24 L

we have below equation to be used:

Molarity,

M = number of mol / volume in L

= 0.36/6.24

= 5.769*10^-2 M

Answer: 5.8*10^-2 M

B)

Molar mass of C2H6O = 2*MM(C) + 6*MM(H) + 1*MM(O)

= 2*12.01 + 6*1.008 + 1*16.0

= 46.068 g/mol

mass of C2H6O = 74.0 g

we have below equation to be used:

number of mol of C2H6O,

n = mass of C2H6O/molar mass of C2H6O

=(74.0 g)/(46.068 g/mol)

= 1.606 mol

volume , V = 2.43 L

we have below equation to be used:

Molarity,

M = number of mol / volume in L

= 1.606/2.43

= 0.661 M

Answer: 0.661 M

C)

Molar mass of KI = 1*MM(K) + 1*MM(I)

= 1*39.1 + 1*126.9

= 166 g/mol

mass of KI = 12.43 mg

= 0.01243 g [using conversion 1 g = 1000 mg]

we have below equation to be used:

number of mol of KI,

n = mass of KI/molar mass of KI

=(0.01243 g)/(166 g/mol)

= 7.488*10^-5 mol

volume , V = 110.3 mL= 0.1103 L

we have below equation to be used:

Molarity,

M = number of mol / volume in L

= 7.488*10^-5/0.1103

= 6.789*10^-4 M

Answer: 6.789*10^-4 M