30) how many grams of ch3oh must be added to water to prepare 210 ml of a solution that is
1.50 m ch3oh?
a) 10.0g
b) 48.0 g
c) 3.20 g
d) 0.210 g
e) 22.08

Mass = 10.0 g

Explanation:

Given data:

Molarity = 1.50 M

Volume = 210 mL ( 0.21 L)

Grams of CH₃OH = ?

Solution:

Number of mole = molarity × volume

Number of mole = 1.50 M × 0.21 L

Number of mole = 0.315 mol

Mass of CH₃OH:

Mass = Number of moles ×  molar mass

Mass = 0.315 mol × 32.04 g/mol

Mass = 10.0 g