Question

You are given NaCl (dry powder), a bottle of 100% methanol (liquid), and distilled water. You need to make 200mL of a solution that is of 1% NaCl in 10% methanol. Explain how you would make this solution.

Answer 1

Reqd. volume of the final solution=200mL

For %solution of (v/v) ratio, it means that 1 vol. of methanol in 9 vol. of distilled water

Then,for 10% methanol, 10 ml of methanol is present in 90 ml of distilled water

for total volume we have to multiply 10 ml of meethanol with 200/90=2.2

Therefore, total vol. of methanol= 10ml*2.2=22ml of methanol

Remaning vol of water=200ml-22ml=178ml

Again, for NaCl 1% in 10% methanol= (1 g of NaCl/100ml water) *10 ml of methanol=0.1 g NaCl

Hence we can prepare the reqd. solution by diluting 22 ml of methanol with 178 ml of distilled water and mixing 0.1 g of NaCl in every 10 ml of methanol.