29.2% (W/W) HCI stock solution has a density of 1.25 g `mL^(-1)`. The molecular mass of HCI is 36.5 g `mol^(-1)` Calculate the volume (mL) of stock solution required to prepare 200 mL of 0.4 M HCI

Options

A. 2 mL
B. 4 mL
C. 8 mL
D. 6 mL

Answer

C

Explanation

Stock solution of HCI = 29.2 % (w/w)
Thus, 29.2 g HCI are present in 100 g of the solution.
As density of solution = 1.25 g `mL^(-1)`,
Volume of 100 g of the solution = 100/1.25 mL
Molar mass of HCI = 36.5 g `mol^(-1)`,
`"Molarity of the solution" = (W_(2) xx 1000)/(M_(2) xx V (cm^(3)))`
`29.2/36.5 xx 1.25/100 xx 1000 = 10 M`
`{:(M_(1) V_(1),=,M_(2) V_(2)),(("stock solution"), ,("solution required")),(10 xx V_(1),=,0.4 xx 200):}`
` or V_(1) = 8 mL`