Question

A chemist has a sample of 1.88 M HCl. What volume of the sample does he require to prepare 200 mL of 0.45 M HCl?

Answer 1

Answer: v = 200 ml · 0.45 M/ 1.88 M

Explanation: when solution is diluted, amount of substance

is same and n= c · V v1 = v2 · c2 / c1

Answer 2

The appropriate answer is "47.87 mL".

Given values are:

• Concentration of undiluted HCl, $C_1=1.88 \ M$
• Concentration of diluted HCl, $C_2=0.45 \ M$
• Volume of diluted HCl, $V_2=200 \ mL$

By using the dilution equation, we get

→ $C_1V_1=C_2V_2$

or,

→     $V_1=\frac{C_2V_2}{C_1}$

By substituting the given values, we get

→          $=\frac{0.45\times 200}{1.88}$

→          $=\frac{90}{1.88}$

→          $= 47.87 \ mL$

Thus the above answer is correct.

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