Question

Describe how you would prepare a liter aqueous solution of each of the following reagents
(a) 1 M glycine
(b) 0.5 M glucose
(c) 10 mM ethanol
(d) 100 mM hemoglobin

Answer 1

Describe how you would prepare a liter aqueous solution of each of the following reagents
(a) 1 M glycine

Ans:

We know that molarity = Moles of solute / volume of solution in litres

So we will need one mole of glycine to make 1M solution of glycine with 1L of volume

Molar mass of glycine = 75g/ mole

mole = mass / molar mass

Mass = Moles X molar mass = 1 X 75 = 75g

We will weigh 75g of glycine and will dissolve it in 1L of solution.

(b) 0.5 M glucose

Ans:

Molarity = moles / volume (L)

moles = Molarity X volume = 0.5 X 1 = 0.5

So we need 0.5 moles of glucose

Molar mass of glucose = 180g/mole

mass = Moles X molar mass = 0.5 X 180 = 90g

So we will weigh 90g of glucose and will dissolve it in 1L of solution

(c) 10 mM ethanol

We need 10mM, so if we start with absolute alcohol (99% pure) then we have 99mL of ethanol in 100mL of solution

density of ethanol = 0.789g / mL

The mass of ethanol in 100mL solution = 99 X 0.789 = 78.11grams

Moles = Mass / molar mass = 78.11/46 = 1.698moles / 100mL

Therefore we have a solution of 16.98 molarity

M1V1 = M2V2

M1 = 16.98 M

V1 = ?

V2 = 1000mL

M2 = 10 X 10^-3 M

V1 = 10 X 10^-3 M X 1000mL / 16.98

V1 = 0.589 mL

so we will dissolve 99% pure ethanol (0.589mL) into total 1000mL of solution

(d) 100 mM hemoglobin

Molar mass of hemoglobin = 64458g/mole

for 100mM we need 100mmoles of hemoglobin in 1L

Mass = Moles X molar mass = 100 X 10^-3 X 64458 = 6445.8 grams

We will dissolve this much mass and make up the solution to 1L [Not possible though]