Given stock solutions of glucose (1m), asparagine (100 mm), and nah2po4 (50 mm), how much of each solution would you need to prepare 500 ml of a reagent that contains 0.05 m glucose, 10 mm asparagine, and 2 mm nah2po4?

25 mL,50 mL, 20 mL

Explanation:

Molarity = numbers of mole / volume in liters

Glucose

stock solutions concentration = 1M

concentration needed = 0.05 M

volume of the preparation = 500ml = 500 / 1000 = 0.5l

number of moles of glucose in the solution = Molarity × molar mass = 0.05 × 0.5 L = 0.025 moles

volume of  the stock needed = number of moles / molarity of the stock solution = 0.025 moles / 1 M = 0.025L = 25 mL

Asparagine

Molarity of the stock solution = 100mM = 0.1 M

Molarity needed = 10mM = 0.01 M

volume of the prepared solution = 0.5 L

number of moles in the prepared solution = 0.01 M × 0.5 L = 0.005 moles

volume of the stock solution needed = number of moles of the prepared solution / Molarity of the stock = 0.005 / 0.1 = 0.05L = 50 mL

NaH₂PO₄

Molarity of the stock solution = 50 mM = 0.05 M

Molarity of the needed = 2 mM = 0.002 M

Volume of the prepared solution = 0.5 L

number of moles in the prepared solution = 0.002 M × 0.5 L = 0.002 × 0.5 L = 0.001 moles

volume of the stock needed = 0.001 moles / 0.05 M = 0.02 L = 20 mL

specific gravity is the density of a substance divided by the density of water. since (at standard temperature and pressure) water has a density of 1 gram/cm3, and since all of the units cancel, specific gravity is usually very close to the same value as density

so "c" is the answer

plz give brainliest