Answer to Question #235957 in General Chemistry for The Decembrist

Question

The solubility of the hydroxide of a hypothetical metal, Me, in water is 144.27 g MeOH per 54.88 mL water. Calculate the amount of water in mL needed to dissolve 32.41 g of MeOH.

Atomic Mass:

Me: 68.32 g/mol

O: 15.999 g/mol

H: 1.00784 g/mol

Answer 1

Solution

MeOH= 68.32+15.99+1.0078= 85.3278

1 mole MeOH is 85.3178g

? Mole MeOH is 144.27

= 1.69 M

If 1.69 moles MeOH is dissolved in 54.88mL of water

How much water will dissolve $\frac {32.41}{85.3178}\times 1= 0.225M$

$\frac{0.225\times54.88}{1.69}= 7.263g/mL$

7.263g/mL of water is required to dissolve 32.41g of MeOH