How many grams of  ${\mathrm{CH}}_3\mathrm{OH}$ should be added to water to prepare

 150 ml solution of  $2{\mathrm{MCH}}_3\mathrm{OH}$

1000 ml of ${\mathrm{CH}}_3\mathrm{OH}$   requires  methanol  = 32 g. 

150 ml of 2 M ${\mathrm{CH}}_3\mathrm{OH}$ requires methanol 

$=\frac{32}{1000}\times 150\times 2=9.6\mathrm{g}$