29.2% (w/w) HCl stock solution has density of 1.25 g/mL.

Question

29.2% (w/w) HCl stock solution has density of 1.25 g/mL. The molecular weight of HCl is 36.5 g/mol. The volume (mL) of stock solution required to prepare a 100 mL solution of 0.4 M HCl is ____

Smart Achievers · Accepted Answer

(4)

$\frac{4 m L}{M} = (\frac{% (w / w) \times d \times 10}{molar mass}) = \frac{(29.2) \times 1.25 \times 10}{36.5} = 10 M$

$According to dilution, equation, M_{i} V_{i} = M_{f} V_{f}$

$V_{f} = \frac{0.4 \times 100}{10} = 4 m L$

Solution

Q.
29.2% (w/w) HCl stock solution has density of 1.25 g/mL. The molecular weight of HCl is 36.5 g/mol. The volume (mL) of stock solution required to prepare a 100 mL solution of 0.4 M HCl is ____

Ans.
(4)

$\frac{4 m L}{M} = (\frac{% (w / w) \times d \times 10}{molar mass}) = \frac{(29.2) \times 1.25 \times 10}{36.5} = 10 M$

$According to dilution, equation, M_{i} V_{i} = M_{f} V_{f}$

$V_{f} = \frac{0.4 \times 100}{10} = 4 m L$

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Solution

Q. 29.2% (w/w) HCl stock solution has density of 1.25 g/mL. The molecular weight of HCl is 36.5 g/mol. The volume (mL) of stock solution required to prepare a 100 mL solution of 0.4 M HCl is ____

Ans. (4) 4mLM=(%(w/w)×d×10molar mass)=(29.2)×1.25×1036.5=10M According to dilution, equation, MiVi=MfVf Vf=0.4×10010=4mL

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Q.
29.2% (w/w) HCl stock solution has density of 1.25 g/mL. The molecular weight of HCl is 36.5 g/mol. The volume (mL) of stock solution required to prepare a 100 mL solution of 0.4 M HCl is ____

Ans.
(4)

$\frac{4 m L}{M} = (\frac{% (w / w) \times d \times 10}{molar mass}) = \frac{(29.2) \times 1.25 \times 10}{36.5} = 10 M$

$According to dilution, equation, M_{i} V_{i} = M_{f} V_{f}$

$V_{f} = \frac{0.4 \times 100}{10} = 4 m L$