Correct Answer - C
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`