1. Calculate the quantity of glucose needed to prepare (a) 250 mL of a 100 mM of solution (in water), (b) 100 mL of 0.05% solution and (c) 500 mL of 50 mg/mL.
[Formula: C6H12O6 Melting point: 146°C, IUPAC ID: D-glucose, Density:
1.56 g/cm, Molar mass: 180.156 g/mol).​

a. mass of glucose required = 4.50 g

b. mass of glucose required = 0.05 g

c. mass of glucose required = 25g

Explanation:

1a. molar mass of glucose = 180.156 g/mol

volume of solution = 250 mL or 250 mL * 1 L/1000 mL = 0.25 L

concentration of solution = 100 mM or 100mM * 1 M/1000 mM = 0.10 M

amount in moles = mass/ molar mass = concentration * volume

mass in grams = concentration * volume * molar mass

mass = 0.10 * 0.25 * 180.156

mass of glucose required = 4.50 g

b. mass concentration of 0.05 % solution = 0.05 g/100 ml = 0.5 g/L

volume of solution = 100 mL = 0.1 L

mass in grams = mass concentration * volume

mass in grams =  0.5 g/L * 0.1 L

mass of glucose required = 0.05 g

c. mass concentration of solution = 50 mg/ mL = 0.05 g/ 0.001 L = 50 g/L

volume of solution = 500 mL = 0.5 L

mass of solute required = concentration * volume

mass required = 50 g/L * 0.5 L

mass of glucose required = 25g

your answer would be a. 12.5 kg