Question

Calculate the amount of benzoic acid required for preparing 250ml of 0.15m solution in methanol

Answer 1

Explanation:

Number of moles of benzoic acid required =0.15 ×

1000

250

=0.0375 moles

Molar mass of benzoic acid =7(12)+6(1)+2(16)=84+6+32=122 g/mol

Mass of benzoic acid required =122×0.0375=4.575 g

Answer 2

Molarity (M) = 0.15M

Volume (V) = 250ml

GMW C$_{6}$H$_{5}$COOH (W) = $\sf\red{ ? }$

• M = $\sf\frac{ W }{ GMW }$ × $\sf\frac{ 1000 }{ V \: (ml) }$

• W = $\sf\frac{ M \: × \: GMW \: V \: (ml) }{ 1000 }$

• W = $\sf\frac{ 0.15 \: × \: 122 \: × \: 250 }{ 1000 }$

• W = 4.575 gm ✓