Question

A) How many grams of HCI are in 300.0 mL of 12.0 M HCI Solution?

B) How many grams of solute are present in 125 mL of 0.20 M NaHCO3?

C) What is the molarity of the solution when 55.0g of glucose C6H12O6 is dissolved in water to make 1250 mL of solution?

D) What is the molarity of the solution when 35.0g of potassium phosphate dissolves in water to make 750. mL of solution?

E) A 0.025 M glucose solution contains 25.0g of glucose. What is the volume of this solution?

F) A 0.42 M Solution of potassium nitrate contains 15.0g of potassium nitrate. What is the volume of this solution?

Answer 1

Solution:

Molarity (M) is defined as:

M = mass / (molar mass x volume in L)

A) How many grams of HCI are in 300.0 mL of 12.0 M HCI Solution?

Volume of HCl = 300 mL = 0.30 L, M = 12 M, molar mass of HCl = 36.46 g/mol

Since, M = mass / (molar mass x volume in L), then

mass = M x molar mass x volume in L = 12 M x 0.30 L x 36.5 g/mol = 240.636 g

B) How many grams of solute are present in 125 mL of 0.20 M NaHCO3?

Volume of NaHCO3 = 125 mL = 0.125 L, M = 0.20 M, molar mass of NaHCO3 = 84.007 g/mol

Since, M = mass / (molar mass x volume in L), then

mass = M x molar mass x volume in L = 0.20 M x 0.125 L x 84.007 g/mol = 2.100 g

C) Molarity of the solution when 55.0 g of glucose C₆H₁₂O₆ is dissolved in water to make 1250 mL of solution?

Volume of C₆H₁₂O₆= 1250 mL = 1.25 L, mass = 55 g, molar mass of C₆H₁₂O₆ = 180.156 g/mol

Since, M = mass / (molar mass x volume in L), then

M= 55 g / 1.25 L x 180.156 g/mol = 0.244 M

D) Molarity of the solution when 35.0g of potassium phosphate dissolves in water to make 750. mL of solution?

Volume of solution = 750 mL = 0.750 L, mass = 35.0 g, molar mass of potassium phosphate = 212.27 g/mol

Since, M = mass / (molar mass x volume in L), then

M= 35 g / 0.750 L x 212.27 g/mol = 0.220 M

E) 0.025 M glucose solution contains 25.0 g of glucose. What is the volume ?

mass = 25 g, molar mass of C₆H₁₂O₆ = 180.156 g/mol, M= 0.025 M

Since, M = mass / (molar mass x volume in L), then

Volume = mass / Molar mass x M

Volume = 25 g / 180.156 g/mol x 0.025 M = 5.550 L = 5550 mL

F) Volume of  0.42 M Solution of potassium nitrate contains 15.0g of potassium nitrate?

mass = 15 g, molar mass of potassium nitrate = 101.1032 g/mol, M= 0.42 M

Since, M = mass / (molar mass x volume in L), then

Volume = mass / Molar mass x M

Volume = 15 g / 101.1032 g/mol x 0.42 M = 0.3532 L = 353.2 mL