Question

Calculate the mass of KI in grams required to prepare 3.00 ✕ 10^​2 mL of a 2.40 M solution.

Calculate the molarity of each of the following solutions. (a) 32.0 g of ethanol (C2H5OH) in 545 mL of solution M

(b) 15.4 g of sucrose (C12H22O11) in 68.0 mL of solution M

(c) 7.00 g of sodium chloride (NaCl) in 86.4 mL of solution M

Answer 1

a)
volume , V = 300 mL
= 0.3 L


number of mol,
n = Molarity * Volume
= 2.4*0.3
= 0.72 mol

Molar mass of KI,
MM = 1*MM(K) + 1*MM(I)
= 1*39.1 + 1*126.9
= 166 g/mol

mass of KI,
m = number of mol * molar mass
= 0.72 mol * 166 g/mol
= 1.20*10^2 g
Answer: 1.20*10^2 g

b)

Molar mass of C12H22O11,
MM = 12*MM(C) + 22*MM(H) + 11*MM(O)
= 12*12.01 + 22*1.008 + 11*16.0
= 342.296 g/mol


mass(C12H22O11)= 15.4 g

number of mol of C12H22O11,
n = mass of C12H22O11/molar mass of C12H22O11
=(15.4 g)/(342.296 g/mol)
= 4.499*10^-2 mol
volume , V = 68.0 mL
= 6.8*10^-2 L


Molarity,
M = number of mol / volume in L
= 4.499*10^-2/6.8*10^-2
= 0.6616 M
AnsweR: 0.662 M

c)
Molar mass of NaCl,
MM = 1*MM(Na) + 1*MM(Cl)
= 1*22.99 + 1*35.45
= 58.44 g/mol


mass(NaCl)= 7.00 g

number of mol of NaCl,
n = mass of NaCl/molar mass of NaCl
=(7.0 g)/(58.44 g/mol)
= 0.1198 mol
volume , V = 86.4 mL
= 8.64*10^-2 L


Molarity,
M = number of mol / volume in L
= 0.1198/8.64*10^-2
= 1.386 M
Answer: 0.139 M