Question

How many grams of ch₃oh must be added to water to prepare 310 ml of a solution that is 4.0 m ch₃oh ?

Answer 1

`\huge{ \boxed{39.68 \: g}}`

Step-by-step explanation:

The mass of
`CH_3OH` required can be found by using the formula:

m = c × M × v

where:

c is the concentration in M , mol/dm³ or mol/L

v is the volume in L or dm³

m is the mass in grams

M is the molar mass in g/mol

From the question:

c = 4.0 M

Molar mass of(M)
`CH_3OH` = 12 + (3 × 1) + 16 + 1 = 12 + 3 + 16 + 1 = 32 g/mol

`1000ml = 1l \\ 310 \: ml = (310)/(1000) * 1l \\ v \: = 0.310 \: l`

`\therefore \: m = 4 * 32 * 0.310 \\ = 39.68 \: g`