29.2 %( w / w ) HCl stock solution has a density of 1.25 g mL -1. The molecular weight of HCl is 36.5 g mol -1 . The volume (in mL ) of stock solution required to prepare a 200 mL solution of 0.4 M HCl is-

Question

29.2 %( w / w ) HCl stock solution has a density of 1.25 g mL -1. The molecular weight of HCl is 36.5 g mol -1 . The volume (in mL ) of stock solution required to prepare a 200 mL solution of 0.4 M HCl is- (A) 16 mL (B) 61 mL (C) 80 mL (D) 8 mL

Tardigrade Admin · Accepted Answer

Density of solution =1.25 g / mL 29.2 %( w / w ) means that 29.2 g of HCl is present in 100 gms of solution So, ρ ( density of solution )=( text wt. of soln. / text vol. of soln. ) 1.25 g / mL =(100/ text vol. of soln ) v =(100/1.25) mL molarity of solution =( text no. of moles of solute / text vol. of solution (in ltrs) ) M =(29.2 / 36.5/100 / 1.25) × 1000=10 M Apply, M 1 V 1= M 2 V 2 0.4 × 200=10 × V V =8 mL

Q. $29.2% (w / w) H Cl$ stock solution has a density of $1.25 g$ $m L^{- 1}$ . The molecular weight of $H Cl$ is $36.5 g m o l^{- 1} .$ The volume (in $m L$ ) of stock solution required to prepare a $200 m L$ solution of $0.4 M H Cl$ is-

$16 m L$

$61 m L$

$80 m L$

$8 m L$

Q. 29.2%(w/w)HCl stock solution has a density of 1.25g mL−1. The molecular weight of HCl is 36.5gmol−1. The volume (in mL ) of stock solution required to prepare a 200mL solution of 0.4MHCl is-

16mL

61mL

80mL

8mL

Q. $29.2% (w / w) H Cl$ stock solution has a density of $1.25 g$ $m L^{- 1}$ . The molecular weight of $H Cl$ is $36.5 g m o l^{- 1} .$ The volume (in $m L$ ) of stock solution required to prepare a $200 m L$ solution of $0.4 M H Cl$ is-

$16 m L$

$61 m L$

$80 m L$

$8 m L$