How many grams of dibasic acid (molecular mass 200) should be present in 100 mL of the aqueous solution to give 0.1 N solution ?

Options

A. 1g
B. 2g
C. 10 g
D. 20 g

Answer

A

Explanation

Molarity of acid `= 0.1//2 = 0.05 M`
Molarity `= (("Mass of acid")/("Molar mass"))/("Volume in litre")`
`(0.05 molL^(-1))=(W)/(("200 g mol"^(-1))xx("0.1 L"))`
`W = (0.05 mol L^(-1))xx("200 g mol"^(-1))xx(0.1 L) =-1g`.