How many grams of CH3OH must be added to water to prepare 275 mL of a solution that is 3.5 M CH3OH?

Question

asked Jul 14, 2024 10.0k views

1 Answer

Ameera · Answer 1 · 2024-07-19T01:40:58+0000

Answer:

30.8 grams

Step-by-step explanation:

n = 3.5 mol/L × 0.275 L = 0.9625 mol

The molar mass of CH3OH is 32 g/mol.

Therefore, the mass of CH3OH required is:

mass = n × molar mass

mass = 0.9625 mol × 32 g/mol = 30.8 g

Therefore, you need 30.8 grams of CH3OH to prepare a solution that is 3.5 M CH3OH in 275 mL of water.