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

Tardigrade Admin · Accepted Answer

Stock solution of HCl =29.2 %( w / w ) Molarity of stock solution of HCl =(29.2 × 1000 × 1.25/36.5 × 100) If volume of stock solution required = V ml V × (29.2/36.5) × (1000/80)=200 × 0.4 ⇒ V =8 ml

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

Stock solution of $H Cl = 29.2% (w / w)$
Molarity of stock solution of $H Cl = \frac{29.2 \times 1000 \times 1.25}{36.5 \times 100}$
If volume of stock solution required $= V m l$
$V \times \frac{29.2}{36.5} \times \frac{1000}{80} = 200 \times 0.4$
$\Rightarrow V = 8 m l$

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

Stock solution of HCl=29.2%(w/w) Molarity of stock solution of HCl=36.5×10029.2×1000×1.25​ If volume of stock solution required =Vml V×36.529.2​×801000​=200×0.4 ⇒V=8ml

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

Stock solution of $H Cl = 29.2% (w / w)$
Molarity of stock solution of $H Cl = \frac{29.2 \times 1000 \times 1.25}{36.5 \times 100}$
If volume of stock solution required $= V m l$
$V \times \frac{29.2}{36.5} \times \frac{1000}{80} = 200 \times 0.4$
$\Rightarrow V = 8 m l$